0=9x^2+20x+5

Solution for 0=9x^2+20x+5 equation:

0=9x^2+20x+5

We move all terms to the left:

0-(9x^2+20x+5)=0

We add all the numbers together, and all the variables

-(9x^2+20x+5)=0

We get rid of parentheses

-9x^2-20x-5=0

a = -9; b = -20; c = -5;
Δ = b2-4ac
Δ = -202-4·(-9)·(-5)
Δ = 220
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:

$\sqrt{\Delta}=\sqrt{220}=\sqrt{4*55}=\sqrt{4}*\sqrt{55}=2\sqrt{55}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{55}}{2*-9}=\frac{20-2\sqrt{55}}{-18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{55}}{2*-9}=\frac{20+2\sqrt{55}}{-18}
$