3x2+36x+12=0

Solution for 3x2+36x+12=0 equation:

3x^2+36x+12=0

a = 3; b = 36; c = +12;
Δ = b2-4ac
Δ = 362-4·3·12
Δ = 1152
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:

$\sqrt{\Delta}=\sqrt{1152}=\sqrt{576*2}=\sqrt{576}*\sqrt{2}=24\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-24\sqrt{2}}{2*3}=\frac{-36-24\sqrt{2}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+24\sqrt{2}}{2*3}=\frac{-36+24\sqrt{2}}{6}
$