m^2=-2m+120

Solution for m^2=-2m+120 equation:

m^2=-2m+120

We move all terms to the left:

m^2-(-2m+120)=0

We get rid of parentheses

m^2+2m-120=0

a = 1; b = 2; c = -120;
Δ = b2-4ac
Δ = 22-4·1·(-120)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-22}{2*1}=\frac{-24}{2}
=-12
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+22}{2*1}=\frac{20}{2}
=10
$