3x^2+36x+9=0

Solution for 3x^2+36x+9=0 equation:

3x^2+36x+9=0

a = 3; b = 36; c = +9;
Δ = b2-4ac
Δ = 362-4·3·9
Δ = 1188
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:

$\sqrt{\Delta}=\sqrt{1188}=\sqrt{36*33}=\sqrt{36}*\sqrt{33}=6\sqrt{33}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-6\sqrt{33}}{2*3}=\frac{-36-6\sqrt{33}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+6\sqrt{33}}{2*3}=\frac{-36+6\sqrt{33}}{6}
$