18=9x^2+4x

Solution for 18=9x^2+4x equation:

18=9x^2+4x

We move all terms to the left:

18-(9x^2+4x)=0

We get rid of parentheses

-9x^2-4x+18=0

a = -9; b = -4; c = +18;
Δ = b2-4ac
Δ = -42-4·(-9)·18
Δ = 664
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:

$\sqrt{\Delta}=\sqrt{664}=\sqrt{4*166}=\sqrt{4}*\sqrt{166}=2\sqrt{166}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{166}}{2*-9}=\frac{4-2\sqrt{166}}{-18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{166}}{2*-9}=\frac{4+2\sqrt{166}}{-18}
$