3x^2+6x-1590=0

Solution for 3x^2+6x-1590=0 equation:

3x^2+6x-1590=0

a = 3; b = 6; c = -1590;
Δ = b2-4ac
Δ = 62-4·3·(-1590)
Δ = 19116
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:

$\sqrt{\Delta}=\sqrt{19116}=\sqrt{324*59}=\sqrt{324}*\sqrt{59}=18\sqrt{59}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-18\sqrt{59}}{2*3}=\frac{-6-18\sqrt{59}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+18\sqrt{59}}{2*3}=\frac{-6+18\sqrt{59}}{6}
$