3x^2+540x+3000=0

Solution for 3x^2+540x+3000=0 equation:

3x^2+540x+3000=0

a = 3; b = 540; c = +3000;
Δ = b2-4ac
Δ = 5402-4·3·3000
Δ = 255600
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:

$\sqrt{\Delta}=\sqrt{255600}=\sqrt{3600*71}=\sqrt{3600}*\sqrt{71}=60\sqrt{71}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(540)-60\sqrt{71}}{2*3}=\frac{-540-60\sqrt{71}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(540)+60\sqrt{71}}{2*3}=\frac{-540+60\sqrt{71}}{6}
$