Solve the following equation I need help to solve the following equation. 
$13x=4x^2$
My attempt: 
$13x=4x^2$
$4x^2-13x=0$
$\frac{4x^2-13x}{4}=0$
$x^2-\frac{13x}{4}+0=0$
$x=-\frac{-13}{2} \begin{matrix} + \\ - \end{matrix} \sqrt{(\frac{-13}{2})^2}$
this results to $x1 = 8.125$ $x2 = -4.875$
The correct result is $x1 = 0$ $x2 = 3.25$
Where did I go wrong with this?
Thanks!!
 A: Nice attempt but you have over complicated things for yourself there!
Let me carry on for from $4x^2-13x=0$
I know that maybe plugging into the quadratic formula may be the first instinct, but we can simply this by factorizing out an $x$. 
$4x^2-13x=4x \times x-13x=x(4x-13)$
So here we get that $x=0$ is a solution which will be your $x_1$.  All that is left to solve is the other bracket $(4x-13)=0$ for your $x_2$.  I think you should get the answer you are looking for!
A: I will go ahead and add this as an answer since it confirms your posted solutions.  
$13x=4x^2$ 
divide both sides by x so that you have 
$13=4x$ 
solve for $x=13/4$ which equals 3.25.  
And of course, 0 exists as the trivial solution.  
A: First you see that $x = 0$ is a solution, because on both sides there are no additional numbers or other factors. Then you assume that x $does not$ equal $0$ to search for more solutions. Then you can just divide by x (this is allowed because x is in this case not 0), which leads to:
$$4x = 13$$
Then you obtain the second solution.
A: $13x=x^2$
If $x\neq 0$ we can divide both sides by $x$ to obtain:
$\frac{13x}{x}=\frac{4x^2}{x}\Rightarrow 13=4x$
Then we divide both sides by $4$ to obtain:
$x = \frac{13}{4}$.
