0=9x^2-20x-4

Solution for 0=9x^2-20x-4 equation:

0=9x^2-20x-4

We move all terms to the left:

0-(9x^2-20x-4)=0

We add all the numbers together, and all the variables

-(9x^2-20x-4)=0

We get rid of parentheses

-9x^2+20x+4=0

a = -9; b = 20; c = +4;
Δ = b2-4ac
Δ = 202-4·(-9)·4
Δ = 544
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:

$\sqrt{\Delta}=\sqrt{544}=\sqrt{16*34}=\sqrt{16}*\sqrt{34}=4\sqrt{34}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{34}}{2*-9}=\frac{-20-4\sqrt{34}}{-18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{34}}{2*-9}=\frac{-20+4\sqrt{34}}{-18}
$