g^2+6g=27

Solution for g^2+6g=27 equation:

g^2+6g=27

We move all terms to the left:

g^2+6g-(27)=0

a = 1; b = 6; c = -27;
Δ = b2-4ac
Δ = 62-4·1·(-27)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-12}{2*1}=\frac{-18}{2}
=-9
$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+12}{2*1}=\frac{6}{2}
=3
$