g2+23g=6

Solution for g2+23g=6 equation:

g2+23g=6

We move all terms to the left:

g2+23g-(6)=0

We add all the numbers together, and all the variables

g^2+23g-6=0

a = 1; b = 23; c = -6;
Δ = b2-4ac
Δ = 232-4·1·(-6)
Δ = 553
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-\sqrt{553}}{2*1}=\frac{-23-\sqrt{553}}{2}
$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+\sqrt{553}}{2*1}=\frac{-23+\sqrt{553}}{2}
$