(8x-4)(12x-20)=36x+8

Solution for (8x-4)(12x-20)=36x+8 equation:

(8x-4)(12x-20)=36x+8

We move all terms to the left:

(8x-4)(12x-20)-(36x+8)=0

We get rid of parentheses

(8x-4)(12x-20)-36x-8=0

We multiply parentheses ..

(+96x^2-160x-48x+80)-36x-8=0

We get rid of parentheses

96x^2-160x-48x-36x+80-8=0

We add all the numbers together, and all the variables

96x^2-244x+72=0

a = 96; b = -244; c = +72;
Δ = b2-4ac
Δ = -2442-4·96·72
Δ = 31888
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{31888}=\sqrt{16*1993}=\sqrt{16}*\sqrt{1993}=4\sqrt{1993}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-244)-4\sqrt{1993}}{2*96}=\frac{244-4\sqrt{1993}}{192}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-244)+4\sqrt{1993}}{2*96}=\frac{244+4\sqrt{1993}}{192}
$