13+24x^2-20x=25x^2+10x

Solution for 13+24x^2-20x=25x^2+10x equation:

13+24x^2-20x=25x^2+10x

We move all terms to the left:

13+24x^2-20x-(25x^2+10x)=0

We get rid of parentheses

24x^2-25x^2-20x-10x+13=0

We add all the numbers together, and all the variables

-1x^2-30x+13=0

a = -1; b = -30; c = +13;
Δ = b2-4ac
Δ = -302-4·(-1)·13
Δ = 952
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{952}=\sqrt{4*238}=\sqrt{4}*\sqrt{238}=2\sqrt{238}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-2\sqrt{238}}{2*-1}=\frac{30-2\sqrt{238}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+2\sqrt{238}}{2*-1}=\frac{30+2\sqrt{238}}{-2}
$