9x^2+2x-396=0

Solution for 9x^2+2x-396=0 equation:

9x^2+2x-396=0

a = 9; b = 2; c = -396;
Δ = b2-4ac
Δ = 22-4·9·(-396)
Δ = 14260
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:

$\sqrt{\Delta}=\sqrt{14260}=\sqrt{4*3565}=\sqrt{4}*\sqrt{3565}=2\sqrt{3565}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{3565}}{2*9}=\frac{-2-2\sqrt{3565}}{18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{3565}}{2*9}=\frac{-2+2\sqrt{3565}}{18}
$