3x(2x+20)+(4x-20)=180

Solution for 3x(2x+20)+(4x-20)=180 equation:

3x(2x+20)+(4x-20)=180

We move all terms to the left:

3x(2x+20)+(4x-20)-(180)=0

We multiply parentheses

6x^2+60x+(4x-20)-180=0

We get rid of parentheses

6x^2+60x+4x-20-180=0

We add all the numbers together, and all the variables

6x^2+64x-200=0

a = 6; b = 64; c = -200;
Δ = b2-4ac
Δ = 642-4·6·(-200)
Δ = 8896
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:

$\sqrt{\Delta}=\sqrt{8896}=\sqrt{64*139}=\sqrt{64}*\sqrt{139}=8\sqrt{139}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(64)-8\sqrt{139}}{2*6}=\frac{-64-8\sqrt{139}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(64)+8\sqrt{139}}{2*6}=\frac{-64+8\sqrt{139}}{12}
$