9x^2+16x-132=0

Solution for 9x^2+16x-132=0 equation:

9x^2+16x-132=0

a = 9; b = 16; c = -132;
Δ = b2-4ac
Δ = 162-4·9·(-132)
Δ = 5008
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:

$\sqrt{\Delta}=\sqrt{5008}=\sqrt{16*313}=\sqrt{16}*\sqrt{313}=4\sqrt{313}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{313}}{2*9}=\frac{-16-4\sqrt{313}}{18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{313}}{2*9}=\frac{-16+4\sqrt{313}}{18}
$