3x^2-36x+20=0

Solution for 3x^2-36x+20=0 equation:

3x^2-36x+20=0

a = 3; b = -36; c = +20;
Δ = b2-4ac
Δ = -362-4·3·20
Δ = 1056
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:

$\sqrt{\Delta}=\sqrt{1056}=\sqrt{16*66}=\sqrt{16}*\sqrt{66}=4\sqrt{66}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-4\sqrt{66}}{2*3}=\frac{36-4\sqrt{66}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+4\sqrt{66}}{2*3}=\frac{36+4\sqrt{66}}{6}
$