3g^2-24g+27=0

Solution for 3g^2-24g+27=0 equation:

3g^2-24g+27=0

a = 3; b = -24; c = +27;
Δ = b2-4ac
Δ = -242-4·3·27
Δ = 252
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{252}=\sqrt{36*7}=\sqrt{36}*\sqrt{7}=6\sqrt{7}$
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-6\sqrt{7}}{2*3}=\frac{24-6\sqrt{7}}{6}
$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+6\sqrt{7}}{2*3}=\frac{24+6\sqrt{7}}{6}
$