3x^2+72x-240=0

Solution for 3x^2+72x-240=0 equation:

3x^2+72x-240=0

a = 3; b = 72; c = -240;
Δ = b2-4ac
Δ = 722-4·3·(-240)
Δ = 8064
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:

$\sqrt{\Delta}=\sqrt{8064}=\sqrt{576*14}=\sqrt{576}*\sqrt{14}=24\sqrt{14}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(72)-24\sqrt{14}}{2*3}=\frac{-72-24\sqrt{14}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(72)+24\sqrt{14}}{2*3}=\frac{-72+24\sqrt{14}}{6}
$