500=9x^2+5x

Solution for 500=9x^2+5x equation:

500=9x^2+5x

We move all terms to the left:

500-(9x^2+5x)=0

We get rid of parentheses

-9x^2-5x+500=0

a = -9; b = -5; c = +500;
Δ = b2-4ac
Δ = -52-4·(-9)·500
Δ = 18025
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:

$\sqrt{\Delta}=\sqrt{18025}=\sqrt{25*721}=\sqrt{25}*\sqrt{721}=5\sqrt{721}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5\sqrt{721}}{2*-9}=\frac{5-5\sqrt{721}}{-18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5\sqrt{721}}{2*-9}=\frac{5+5\sqrt{721}}{-18}
$