g2-22g+120=0

Solution for g2-22g+120=0 equation:

g2-22g+120=0

We add all the numbers together, and all the variables

g^2-22g+120=0

a = 1; b = -22; c = +120;
Δ = b2-4ac
Δ = -222-4·1·120
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2}{2*1}=\frac{20}{2}
=10
$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2}{2*1}=\frac{24}{2}
=12
$