3x^2+27x=39

Solution for 3x^2+27x=39 equation:

3x^2+27x=39

We move all terms to the left:

3x^2+27x-(39)=0

a = 3; b = 27; c = -39;
Δ = b2-4ac
Δ = 272-4·3·(-39)
Δ = 1197
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1197}=\sqrt{9*133}=\sqrt{9}*\sqrt{133}=3\sqrt{133}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-3\sqrt{133}}{2*3}=\frac{-27-3\sqrt{133}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+3\sqrt{133}}{2*3}=\frac{-27+3\sqrt{133}}{6}
$