g^2+g=110

Solution for g^2+g=110 equation:

g^2+g=110

We move all terms to the left:

g^2+g-(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:
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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