m^2=10m+11

Solution for m^2=10m+11 equation:

m^2=10m+11

We move all terms to the left:

m^2-(10m+11)=0

We get rid of parentheses

m^2-10m-11=0

a = 1; b = -10; c = -11;
Δ = b2-4ac
Δ = -102-4·1·(-11)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-12}{2*1}=\frac{-2}{2}
=-1
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+12}{2*1}=\frac{22}{2}
=11
$