m2+22m+120=0

Solution for m2+22m+120=0 equation:

m2+22m+120=0

We add all the numbers together, and all the variables

m^2+22m+120=0

a = 1; b = 22; c = +120;
Δ = b2-4ac
Δ = 222-4·1·120
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2}{2*1}=\frac{-24}{2}
=-12
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2}{2*1}=\frac{-20}{2}
=-10
$