3x^2+6x-960=0

Solution for 3x^2+6x-960=0 equation:

3x^2+6x-960=0

a = 3; b = 6; c = -960;
Δ = b2-4ac
Δ = 62-4·3·(-960)
Δ = 11556
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:

$\sqrt{\Delta}=\sqrt{11556}=\sqrt{36*321}=\sqrt{36}*\sqrt{321}=6\sqrt{321}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6\sqrt{321}}{2*3}=\frac{-6-6\sqrt{321}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{321}}{2*3}=\frac{-6+6\sqrt{321}}{6}
$