(2x)(3x-20)=180

Solution for (2x)(3x-20)=180 equation:

(2x)(3x-20)=180

We move all terms to the left:

(2x)(3x-20)-(180)=0

We multiply parentheses

6x^2-40x-180=0

a = 6; b = -40; c = -180;
Δ = b2-4ac
Δ = -402-4·6·(-180)
Δ = 5920
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:

$\sqrt{\Delta}=\sqrt{5920}=\sqrt{16*370}=\sqrt{16}*\sqrt{370}=4\sqrt{370}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-4\sqrt{370}}{2*6}=\frac{40-4\sqrt{370}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+4\sqrt{370}}{2*6}=\frac{40+4\sqrt{370}}{12}
$