3x^2+36x+4=0

Solution for 3x^2+36x+4=0 equation:

3x^2+36x+4=0

a = 3; b = 36; c = +4;
Δ = b2-4ac
Δ = 362-4·3·4
Δ = 1248
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:

$\sqrt{\Delta}=\sqrt{1248}=\sqrt{16*78}=\sqrt{16}*\sqrt{78}=4\sqrt{78}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-4\sqrt{78}}{2*3}=\frac{-36-4\sqrt{78}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+4\sqrt{78}}{2*3}=\frac{-36+4\sqrt{78}}{6}
$