3x^2+39x-108=0

Solution for 3x^2+39x-108=0 equation:

3x^2+39x-108=0

a = 3; b = 39; c = -108;
Δ = b2-4ac
Δ = 392-4·3·(-108)
Δ = 2817
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:

$\sqrt{\Delta}=\sqrt{2817}=\sqrt{9*313}=\sqrt{9}*\sqrt{313}=3\sqrt{313}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(39)-3\sqrt{313}}{2*3}=\frac{-39-3\sqrt{313}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+3\sqrt{313}}{2*3}=\frac{-39+3\sqrt{313}}{6}
$