(x+20)(3x+60)=180

Solution for (x+20)(3x+60)=180 equation:

(x+20)(3x+60)=180

We move all terms to the left:

(x+20)(3x+60)-(180)=0

We multiply parentheses ..

(+3x^2+60x+60x+1200)-180=0

We get rid of parentheses

3x^2+60x+60x+1200-180=0

We add all the numbers together, and all the variables

3x^2+120x+1020=0

a = 3; b = 120; c = +1020;
Δ = b2-4ac
Δ = 1202-4·3·1020
Δ = 2160
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:

$\sqrt{\Delta}=\sqrt{2160}=\sqrt{144*15}=\sqrt{144}*\sqrt{15}=12\sqrt{15}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(120)-12\sqrt{15}}{2*3}=\frac{-120-12\sqrt{15}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(120)+12\sqrt{15}}{2*3}=\frac{-120+12\sqrt{15}}{6}
$