(11+x)(400-20x)=0

Solution for (11+x)(400-20x)=0 equation:

(11+x)(400-20x)=0

We add all the numbers together, and all the variables

(x+11)(-20x+400)=0

We multiply parentheses ..

(-20x^2+400x-220x+4400)=0

We get rid of parentheses

-20x^2+400x-220x+4400=0

We add all the numbers together, and all the variables

-20x^2+180x+4400=0

a = -20; b = 180; c = +4400;
Δ = b2-4ac
Δ = 1802-4·(-20)·4400
Δ = 384400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{384400}=620$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(180)-620}{2*-20}=\frac{-800}{-40}
=+20
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(180)+620}{2*-20}=\frac{440}{-40}
=-11
$