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 = 1344
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:

$\sqrt{\Delta}=\sqrt{1344}=\sqrt{64*21}=\sqrt{64}*\sqrt{21}=8\sqrt{21}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-8\sqrt{21}}{2*3}=\frac{36-8\sqrt{21}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+8\sqrt{21}}{2*3}=\frac{36+8\sqrt{21}}{6}
$