m^2-13=-10m

Solution for m^2-13=-10m equation:

m^2-13=-10m

We move all terms to the left:

m^2-13-(-10m)=0

We get rid of parentheses

m^2+10m-13=0

a = 1; b = 10; c = -13;
Δ = b2-4ac
Δ = 102-4·1·(-13)
Δ = 152
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:

$\sqrt{\Delta}=\sqrt{152}=\sqrt{4*38}=\sqrt{4}*\sqrt{38}=2\sqrt{38}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{38}}{2*1}=\frac{-10-2\sqrt{38}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{38}}{2*1}=\frac{-10+2\sqrt{38}}{2}
$