(2b-2)(b+11)=0

Solution for (2b-2)(b+11)=0 equation:

(2b-2)(b+11)=0

We multiply parentheses ..

(+2b^2+22b-2b-22)=0

We get rid of parentheses

2b^2+22b-2b-22=0

We add all the numbers together, and all the variables

2b^2+20b-22=0

a = 2; b = 20; c = -22;
Δ = b2-4ac
Δ = 202-4·2·(-22)
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-24}{2*2}=\frac{-44}{4}
=-11
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+24}{2*2}=\frac{4}{4}
=1
$