m^2=-2m+48

Solution for m^2=-2m+48 equation:

m^2=-2m+48

We move all terms to the left:

m^2-(-2m+48)=0

We get rid of parentheses

m^2+2m-48=0

a = 1; b = 2; c = -48;
Δ = b2-4ac
Δ = 22-4·1·(-48)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-14}{2*1}=\frac{-16}{2}
=-8
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+14}{2*1}=\frac{12}{2}
=6
$