4x^2=8x+96

Solution for 4x^2=8x+96 equation:

4x^2=8x+96

We move all terms to the left:

4x^2-(8x+96)=0

We get rid of parentheses

4x^2-8x-96=0

a = 4; b = -8; c = -96;
Δ = b2-4ac
Δ = -82-4·4·(-96)
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-40}{2*4}=\frac{-32}{8}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+40}{2*4}=\frac{48}{8}
=6
$