4x^2-64=12x+8

Solution for 4x^2-64=12x+8 equation:

4x^2-64=12x+8

We move all terms to the left:

4x^2-64-(12x+8)=0

We get rid of parentheses

4x^2-12x-8-64=0

We add all the numbers together, and all the variables

4x^2-12x-72=0

a = 4; b = -12; c = -72;
Δ = b2-4ac
Δ = -122-4·4·(-72)
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-36}{2*4}=\frac{-24}{8}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+36}{2*4}=\frac{48}{8}
=6
$