(10x-20)(6x+8)=12

Solution for (10x-20)(6x+8)=12 equation:

(10x-20)(6x+8)=12

We move all terms to the left:

(10x-20)(6x+8)-(12)=0

We multiply parentheses ..

(+60x^2+80x-120x-160)-12=0

We get rid of parentheses

60x^2+80x-120x-160-12=0

We add all the numbers together, and all the variables

60x^2-40x-172=0

a = 60; b = -40; c = -172;
Δ = b2-4ac
Δ = -402-4·60·(-172)
Δ = 42880
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:

$\sqrt{\Delta}=\sqrt{42880}=\sqrt{64*670}=\sqrt{64}*\sqrt{670}=8\sqrt{670}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-8\sqrt{670}}{2*60}=\frac{40-8\sqrt{670}}{120}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+8\sqrt{670}}{2*60}=\frac{40+8\sqrt{670}}{120}
$