-3x^2+39x=108

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

-3x^2+39x=108

We move all terms to the left:

-3x^2+39x-(108)=0

a = -3; b = 39; c = -108;
Δ = b2-4ac
Δ = 392-4·(-3)·(-108)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(39)-15}{2*-3}=\frac{-54}{-6}
=+9
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+15}{2*-3}=\frac{-24}{-6}
=+4
$