3x^2+3x-99=0

Solution for 3x^2+3x-99=0 equation:

3x^2+3x-99=0

a = 3; b = 3; c = -99;
Δ = b2-4ac
Δ = 32-4·3·(-99)
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{-(3)-3\sqrt{133}}{2*3}=\frac{-3-3\sqrt{133}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{133}}{2*3}=\frac{-3+3\sqrt{133}}{6}
$