m^2-15m+48=-8

Solution for m^2-15m+48=-8 equation:

m^2-15m+48=-8

We move all terms to the left:

m^2-15m+48-(-8)=0

We add all the numbers together, and all the variables

m^2-15m+56=0

a = 1; b = -15; c = +56;
Δ = b2-4ac
Δ = -152-4·1·56
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-1}{2*1}=\frac{14}{2}
=7
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+1}{2*1}=\frac{16}{2}
=8
$