3x^2+6x-640=0

Solution for 3x^2+6x-640=0 equation:

3x^2+6x-640=0

a = 3; b = 6; c = -640;
Δ = b2-4ac
Δ = 62-4·3·(-640)
Δ = 7716
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:

$\sqrt{\Delta}=\sqrt{7716}=\sqrt{4*1929}=\sqrt{4}*\sqrt{1929}=2\sqrt{1929}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{1929}}{2*3}=\frac{-6-2\sqrt{1929}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{1929}}{2*3}=\frac{-6+2\sqrt{1929}}{6}
$