m^2=-10m+15

Solution for m^2=-10m+15 equation:

m^2=-10m+15

We move all terms to the left:

m^2-(-10m+15)=0

We get rid of parentheses

m^2+10m-15=0

a = 1; b = 10; c = -15;
Δ = b2-4ac
Δ = 102-4·1·(-15)
Δ = 160
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-4\sqrt{10}}{2*1}=\frac{-10-4\sqrt{10}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+4\sqrt{10}}{2*1}=\frac{-10+4\sqrt{10}}{2}
$