2x^2-20x+96=64

Solution for 2x^2-20x+96=64 equation:

2x^2-20x+96=64

We move all terms to the left:

2x^2-20x+96-(64)=0

We add all the numbers together, and all the variables

2x^2-20x+32=0

a = 2; b = -20; c = +32;
Δ = b2-4ac
Δ = -202-4·2·32
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-12}{2*2}=\frac{8}{4}
=2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+12}{2*2}=\frac{32}{4}
=8
$