4x^2-99=5x^2+20x

Solution for 4x^2-99=5x^2+20x equation:

4x^2-99=5x^2+20x

We move all terms to the left:

4x^2-99-(5x^2+20x)=0

We get rid of parentheses

4x^2-5x^2-20x-99=0

We add all the numbers together, and all the variables

-1x^2-20x-99=0

a = -1; b = -20; c = -99;
Δ = b2-4ac
Δ = -202-4·(-1)·(-99)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2}{2*-1}=\frac{18}{-2}
=-9
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2}{2*-1}=\frac{22}{-2}
=-11
$