(4x-8)(5x+6)=20x+4x-12

Solution for (4x-8)(5x+6)=20x+4x-12 equation:

(4x-8)(5x+6)=20x+4x-12

We move all terms to the left:

(4x-8)(5x+6)-(20x+4x-12)=0

We add all the numbers together, and all the variables

(4x-8)(5x+6)-(24x-12)=0

We get rid of parentheses

(4x-8)(5x+6)-24x+12=0

We multiply parentheses ..

(+20x^2+24x-40x-48)-24x+12=0

We get rid of parentheses

20x^2+24x-40x-24x-48+12=0

We add all the numbers together, and all the variables

20x^2-40x-36=0

a = 20; b = -40; c = -36;
Δ = b2-4ac
Δ = -402-4·20·(-36)
Δ = 4480
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:

$\sqrt{\Delta}=\sqrt{4480}=\sqrt{64*70}=\sqrt{64}*\sqrt{70}=8\sqrt{70}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-8\sqrt{70}}{2*20}=\frac{40-8\sqrt{70}}{40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+8\sqrt{70}}{2*20}=\frac{40+8\sqrt{70}}{40}
$