2x^2+6x-405=0

Solution for 2x^2+6x-405=0 equation:

2x^2+6x-405=0

a = 2; b = 6; c = -405;
Δ = b2-4ac
Δ = 62-4·2·(-405)
Δ = 3276
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:

$\sqrt{\Delta}=\sqrt{3276}=\sqrt{36*91}=\sqrt{36}*\sqrt{91}=6\sqrt{91}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6\sqrt{91}}{2*2}=\frac{-6-6\sqrt{91}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{91}}{2*2}=\frac{-6+6\sqrt{91}}{4}
$