(20x-10)(11x+4)=180

Solution for (20x-10)(11x+4)=180 equation:

(20x-10)(11x+4)=180

We move all terms to the left:

(20x-10)(11x+4)-(180)=0

We multiply parentheses ..

(+220x^2+80x-110x-40)-180=0

We get rid of parentheses

220x^2+80x-110x-40-180=0

We add all the numbers together, and all the variables

220x^2-30x-220=0

a = 220; b = -30; c = -220;
Δ = b2-4ac
Δ = -302-4·220·(-220)
Δ = 194500
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:

$\sqrt{\Delta}=\sqrt{194500}=\sqrt{100*1945}=\sqrt{100}*\sqrt{1945}=10\sqrt{1945}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-10\sqrt{1945}}{2*220}=\frac{30-10\sqrt{1945}}{440}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+10\sqrt{1945}}{2*220}=\frac{30+10\sqrt{1945}}{440}
$