3x^2+39x-36=0

Solution for 3x^2+39x-36=0 equation:

3x^2+39x-36=0

a = 3; b = 39; c = -36;
Δ = b2-4ac
Δ = 392-4·3·(-36)
Δ = 1953
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:

$\sqrt{\Delta}=\sqrt{1953}=\sqrt{9*217}=\sqrt{9}*\sqrt{217}=3\sqrt{217}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(39)-3\sqrt{217}}{2*3}=\frac{-39-3\sqrt{217}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+3\sqrt{217}}{2*3}=\frac{-39+3\sqrt{217}}{6}
$