x2-4x=16x+96

Solution for x2-4x=16x+96 equation:

x2-4x=16x+96

We move all terms to the left:

x2-4x-(16x+96)=0

We add all the numbers together, and all the variables

x^2-4x-(16x+96)=0

We get rid of parentheses

x^2-4x-16x-96=0

We add all the numbers together, and all the variables

x^2-20x-96=0

a = 1; b = -20; c = -96;
Δ = b2-4ac
Δ = -202-4·1·(-96)
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-28}{2*1}=\frac{-8}{2}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+28}{2*1}=\frac{48}{2}
=24
$