3x^2+12x-99=0

Solution for 3x^2+12x-99=0 equation:

3x^2+12x-99=0

a = 3; b = 12; c = -99;
Δ = b2-4ac
Δ = 122-4·3·(-99)
Δ = 1332
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:

$\sqrt{\Delta}=\sqrt{1332}=\sqrt{36*37}=\sqrt{36}*\sqrt{37}=6\sqrt{37}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-6\sqrt{37}}{2*3}=\frac{-12-6\sqrt{37}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+6\sqrt{37}}{2*3}=\frac{-12+6\sqrt{37}}{6}
$