(r+11)(r-1)=0

Solution for (r+11)(r-1)=0 equation:

(r+11)(r-1)=0

We multiply parentheses ..

(+r^2-1r+11r-11)=0

We get rid of parentheses

r^2-1r+11r-11=0

We add all the numbers together, and all the variables

r^2+10r-11=0

a = 1; b = 10; c = -11;
Δ = b2-4ac
Δ = 102-4·1·(-11)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-12}{2*1}=\frac{-22}{2}
=-11
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+12}{2*1}=\frac{2}{2}
=1
$