8x^2+96=64x

Solution for 8x^2+96=64x equation:

8x^2+96=64x

We move all terms to the left:

8x^2+96-(64x)=0

a = 8; b = -64; c = +96;
Δ = b2-4ac
Δ = -642-4·8·96
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-64)-32}{2*8}=\frac{32}{16}
=2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-64)+32}{2*8}=\frac{96}{16}
=6
$