3x^2+36x+11=0

Solution for 3x^2+36x+11=0 equation:

3x^2+36x+11=0

a = 3; b = 36; c = +11;
Δ = b2-4ac
Δ = 362-4·3·11
Δ = 1164
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:

$\sqrt{\Delta}=\sqrt{1164}=\sqrt{4*291}=\sqrt{4}*\sqrt{291}=2\sqrt{291}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-2\sqrt{291}}{2*3}=\frac{-36-2\sqrt{291}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+2\sqrt{291}}{2*3}=\frac{-36+2\sqrt{291}}{6}
$