3x^2-36x+99=0

Solution for 3x^2-36x+99=0 equation:

3x^2-36x+99=0

a = 3; b = -36; c = +99;
Δ = b2-4ac
Δ = -362-4·3·99
Δ = 108
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:

$\sqrt{\Delta}=\sqrt{108}=\sqrt{36*3}=\sqrt{36}*\sqrt{3}=6\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-6\sqrt{3}}{2*3}=\frac{36-6\sqrt{3}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+6\sqrt{3}}{2*3}=\frac{36+6\sqrt{3}}{6}
$