0=9x^2-22x-32

Solution for 0=9x^2-22x-32 equation:

0=9x^2-22x-32

We move all terms to the left:

0-(9x^2-22x-32)=0

We add all the numbers together, and all the variables

-(9x^2-22x-32)=0

We get rid of parentheses

-9x^2+22x+32=0

a = -9; b = 22; c = +32;
Δ = b2-4ac
Δ = 222-4·(-9)·32
Δ = 1636
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:

$\sqrt{\Delta}=\sqrt{1636}=\sqrt{4*409}=\sqrt{4}*\sqrt{409}=2\sqrt{409}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{409}}{2*-9}=\frac{-22-2\sqrt{409}}{-18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{409}}{2*-9}=\frac{-22+2\sqrt{409}}{-18}
$