(4x-20)(3x+2)=137

Solution for (4x-20)(3x+2)=137 equation:

(4x-20)(3x+2)=137

We move all terms to the left:

(4x-20)(3x+2)-(137)=0

We multiply parentheses ..

(+12x^2+8x-60x-40)-137=0

We get rid of parentheses

12x^2+8x-60x-40-137=0

We add all the numbers together, and all the variables

12x^2-52x-177=0

a = 12; b = -52; c = -177;
Δ = b2-4ac
Δ = -522-4·12·(-177)
Δ = 11200
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:

$\sqrt{\Delta}=\sqrt{11200}=\sqrt{1600*7}=\sqrt{1600}*\sqrt{7}=40\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-52)-40\sqrt{7}}{2*12}=\frac{52-40\sqrt{7}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-52)+40\sqrt{7}}{2*12}=\frac{52+40\sqrt{7}}{24}
$