3x^2+2x-240=0

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

3x^2+2x-240=0

a = 3; b = 2; c = -240;
Δ = b2-4ac
Δ = 22-4·3·(-240)
Δ = 2884
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:

$\sqrt{\Delta}=\sqrt{2884}=\sqrt{4*721}=\sqrt{4}*\sqrt{721}=2\sqrt{721}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{721}}{2*3}=\frac{-2-2\sqrt{721}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{721}}{2*3}=\frac{-2+2\sqrt{721}}{6}
$