3x^2-6x-540=0

Solution for 3x^2-6x-540=0 equation:

3x^2-6x-540=0

a = 3; b = -6; c = -540;
Δ = b2-4ac
Δ = -62-4·3·(-540)
Δ = 6516
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:

$\sqrt{\Delta}=\sqrt{6516}=\sqrt{36*181}=\sqrt{36}*\sqrt{181}=6\sqrt{181}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6\sqrt{181}}{2*3}=\frac{6-6\sqrt{181}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6\sqrt{181}}{2*3}=\frac{6+6\sqrt{181}}{6}
$