I know that to solve $$ax^2+bx+c=0$$  you have to use the formula $$x=  \frac{-b \pm \sqrt{(b^2-4ac)}}{2a}$$  What about more complex ones, like $ax^3+bx^2+cx+d=0$? And what about representing it on a cordinate plane, $2D$ or $3D$?
Just like if we have any quadratic equation which has complex roots, then we are not able to factorize it easily. So we apply quadratic formula and get the roots.  Similarly if we have a cubic equation which has two complex roots (which we know conjugate of each other) and one fractional root, then we are not able to find its first root by hit &amp; trial.  So my question is like quadratic formula, is there exist any thing like cubic formula which help in solving cubic equations?  For example, I have an equation $$2x^3+9x^2+9x-7=0\tag{1}$$ and I have to find its solution which I am not able to find because it has no integral solution. Its solutions are $\dfrac {1}{2}$, $\dfrac{-5\pm \sqrt{3}i}{2} $, I know these solutions because this equation is generated by myself.  So how can I solve equations like these?  Also while typing this question, I thought about the derivation of quadratic formula, which is derived by completing the square method.  So I tried to apply ‘completing the cube’ method on the general equation $ax^3+bx^2+cx+d=0$ but it didn't help.  So please help me in finding a cubic formula or to solve the equations like given in example by an alternative method.
Eclipse says: “Workspace in use or cannot be created, chose a different one.” How do I unlock a workspace? Do extension fields always belong to a bigger field? Let $F$ be a field, $E_1$ and $E_2$ are two distinct extension fields of $F$. Is it the case that we can always somehow find a field $G$ that contains both $E_1$ and $E_2$? In other words, could extensions of fields have different 'direction's such that they are incompatible?  Edit: I began to think about this problem while reading a proof. $F$ is a field. $a$ and $b$ are algebraic over $F$. $p(x)$ and $q(x)$ are two polynomials in $F[x]$ of minimum degree that respectively make $a$ and $b$ a zero. The proof claims that there is an extension $K$ of $F$ such that all distinct zeros of $p(x)$ and $q(x)$ lie in $K$. For a single polynomial, I know this kind of field exists because of the existence of splitting field, why it is true for two polynomials? Pointwise convergence of $h_n(x) = x^{1+\frac{1}{2n-1}}$ Okay, maybe I'm missing something obvious but consider $h_n(x) = x^{ 1+ \frac{1}{2n-1}}$ on the set $[-1,1]$. Now consider a fixed $x \in [-1,1]$, we have:  $$\lim_{n\rightarrow \infty} h_n(x) = x \lim_{n \rightarrow \infty} x^\frac{1}{2n-1} = x$$  But my book says it's $|x|$? Why? In the  question one of the answers had the following code snippet:  IEnumerable&lt;object&gt; FilteredList() {     foreach(object item in FullList)     {         if(IsItemInPartialList(item))             yield return item;     } }   What does the yield keyword do there?  I've seen it referenced in a couple places, and one other question, but I haven't quite figured out what it actually does.  I'm used to thinking of yield in the sense of one thread yielding to another, but that doesn't seem relevant here. I have encountered this question while tring to figure out why every sequential space $X$, is a quotient spaces of a metric space. If I understand correctly, given a sequential space, every sequence (including it's limit) can be identified with the space $Y=\{0\} \cup \{\frac1{n+1}|n \in N\}$. So The space $X$ can viewed as a quotient space of $\bigoplus_{(x_n) \in C} \{(x_n) \} \times Y$ where $C$ is the set of all converges sequences in $X$. My question is, How can I show that this space is metrizable?  Thank you! Shir Is the double dash still the posix standard for long switches? Why multiplicative group $\mathbb{Z}_n^*$ is not cyclic for $n = 2^k$ and $k \ge 3$ How to show hidden folders in File Manager (Nautilus) on Ubuntu? Arduino only works in root I've installed Arduino but as the title says, I can only upload with sudo. And yes my user is part of the dialout group so that's the weird thing.   Arduino did function before on my system. What could be wrong?  Output ls -l:  ls -l /dev/ttyUSB* crw-rw-rw- 1 root dialout 188, 0 apr  2 09:23 /dev/ttyUSB0 Specify a field in attribute table with raster calculator I am trying to use the Alt+Right Click to select edge loops in Linux Mint but this shortcut doesn't work it is already used for Linux to reduce a window. I have checked the keyboard shortcut but it doesn't appear. iPhone App › Add voice recognition? biblatex ieee style same authors Installing grounded light fixture on box with no ground conductor I'm installing a brand new motion activated security light to an existing box on my garage. After removing the old light, I notice it only has white/black wiring while the new light fixture has of course white/black/green (ground) wires to attach. What do I do if no ground wire is available with the old box? Existence of a local geodesic frame Let $(M,g)$ be a Riemannian manifold of dimension $n$ with Riemannian connection $\nabla,$ and let $p \in M.$ Show that there exists a neighborhood $U \subset M$ of $p$ and $n$ (smooth) vector fields $E_1,...,E_n \in \chi (U),$ orthonormal at each point of $U,$ s.t. at $p,$ $\nabla_{E_i}E_j(p)=0.$  What I've found so far;  Let $V$ be a normal neighborhood at $p.$ Let $(W,x(x_1,...,x_n))$ be a local coordinate system at $p$ s.t. $x(W) \subset V.$  Let $\{X_i:=\frac{\partial}{\partial x_i}\}$ be the standard basis of $T_pM.$ By Gram-Schmidt, there exists an orthonormal basis $\{E_i\}$ of $T_pM.$ Consider the orthonormal frame $E_i(t),$ by parallel transporting $\{E_i(0)=E_i\}$ along a  a geodesic $\gamma :I \to M$ starting at $\gamma(0)=p$ and ending in $\gamma(1)=q,$ where $q \in V.$ This gives us an orthonormal frame at each point of $V.$ To satisfy the second condition, let $E_i(t)=\sum_{l}a_{li}X_l$ and  $E_j(t)=\sum_{s}b_{sj}(t)X_s.$ Then   $$\nabla_{E_i}E_j=\sum_{k}(\sum_{l,s} a_{li}(t)b_{sj}(t)\Gamma^k_{ls}+E_i(b_{kj}))X_k,$$  where $\Gamma^k_{is}$ are Christoffel symbols.   To have $\nabla_{E_i}E_j(p)=0,$ we must have $\sum_{l,s}a_{li}(0)b_{sj}(0)\Gamma^k_{ls}+E_i(b_{kj})(p)=0$ for each $k.$      How can I conclude the argument? In chapter 1, of Landau-Lifshitz Mechanics' book, Landau through isotropy and homogeneity of space and homogeneity of time proves that the Lagrangian must depend of magnitude of velocity of the particle.  This seemed fine. But he didn't give a reason as to why it must depend of $|v|^2$ and not on some $|v|^n$ where $n\neq2$. How to create a fancy table of contents for Beamer How can I create a fancy-looking table of contents for a Beamer presentation?    \documentclass{beamer} \usepackage[utf8]{inputenc} \usepackage[T1]{fontenc}  \begin{document}  \frame{ \tableofcontents }  \section{Foo bar} \frame{} \section{Foo bar} \frame{} \section{Foo bar} \frame{} \section{Foo bar} \frame{} \section{Foo bar} \frame{} \section{Foo bar} \frame{} \section{Foo bar} \frame{}  \end{document} How to use an optimization solver to get t-stats and p-values for the estimates? Why isn't it $E \approx 27.642 \times mc^2$? Why is it that many of the most important physical equations don't have ugly numbers (i.e., "arbitrary" irrational factors) to line up both sides?   Why can so many equations be expressed so neatly with small natural numbers while recycling a relatively small set of physical and mathematical constants?  For example, why is mass–energy equivalence describable by the equation $E = mc^2$ and not something like $E \approx 27.642 \times mc^2$?  Why is time dilation describable by something as neat as $t' = \sqrt{\frac{t}{1 - \frac{v^2}{c^2}}}$ and not something ugly like $t' \approx 672.097 \times 10^{-4} \times \sqrt{\frac{t}{1 - \frac{v^2}{c^2}}}$.  ... and so forth.    I'm not well educated on matters of physics and so I feel a bit sheepish asking this.   Likewise I'm not sure if this is a more philosophical question or one that permits a concrete answer ... or perhaps even the premise of the question itself is flawed ... so I would gratefully consider anything that sheds light on the nature of the question itself as an answer.    EDIT: I just wished to give a little more context as to where I was coming from with this question based on some of the responses:  @Jerry Schirmer comments:      You do have an ugly factor of $2.997458 \times 10^8 m/s$ in front of everything. You just hide the ugliness by calling this number c.   These are not the types of "ugly constants" I'm talking about in that this number is the speed of light. It is not just some constant needed to balance two side of an equation.  @Carl Witthoft answers:      It's all in how you define the units ...   Of course this is true, we could in theory hide all sorts of ugly constants by using different units on the right and the left. But as in the case of $E=mc^2$, I am talking about equations where the units on the left are consistent with the units on the right, irrespective of the units used. As I mentioned on a comment there:     $E=mc^2$  could be defined using units like $m$ in imperial stones ($\textsf{S}$), $c$ in cubits/fortnight ($\textsf{CF}^{−1}$) and $E$ in ... umm ... $\textsf{SC}^2\textsf{F}^{−2}$ ... so long as the units are in the same system, we still don't need a fudge factor.   When the units are consistent in this manner, there's no room for hiding fudge constants. Pdf of the square of a standard normal random variable