0=m^2-110+m

Solution for 0=m^2-110+m equation:

0=m^2-110+m

We move all terms to the left:

0-(m^2-110+m)=0

We add all the numbers together, and all the variables

-(m^2-110+m)=0

We get rid of parentheses

-m^2-m+110=0

We add all the numbers together, and all the variables

-1m^2-1m+110=0

a = -1; b = -1; c = +110;
Δ = b2-4ac
Δ = -12-4·(-1)·110
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-21}{2*-1}=\frac{-20}{-2}
=+10
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+21}{2*-1}=\frac{22}{-2}
=-11
$