(11+m)(2+m)=0

Solution for (11+m)(2+m)=0 equation:

(11+m)(2+m)=0

We add all the numbers together, and all the variables

(m+11)(m+2)=0

We multiply parentheses ..

(+m^2+2m+11m+22)=0

We get rid of parentheses

m^2+2m+11m+22=0

We add all the numbers together, and all the variables

m^2+13m+22=0

a = 1; b = 13; c = +22;
Δ = b2-4ac
Δ = 132-4·1·22
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-9}{2*1}=\frac{-22}{2}
=-11
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+9}{2*1}=\frac{-4}{2}
=-2
$