(2x+20)(3x+10)=180

Solution for (2x+20)(3x+10)=180 equation:

(2x+20)(3x+10)=180

We move all terms to the left:

(2x+20)(3x+10)-(180)=0

We multiply parentheses ..

(+6x^2+20x+60x+200)-180=0

We get rid of parentheses

6x^2+20x+60x+200-180=0

We add all the numbers together, and all the variables

6x^2+80x+20=0

a = 6; b = 80; c = +20;
Δ = b2-4ac
Δ = 802-4·6·20
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{-(80)-4\sqrt{370}}{2*6}=\frac{-80-4\sqrt{370}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+4\sqrt{370}}{2*6}=\frac{-80+4\sqrt{370}}{12}
$