18x^2-9x=03

Solution for 18x^2-9x=03 equation:

18x^2-9x=03

We move all terms to the left:

18x^2-9x-(03)=0

We add all the numbers together, and all the variables

18x^2-9x-3=0

a = 18; b = -9; c = -3;
Δ = b2-4ac
Δ = -92-4·18·(-3)
Δ = 297
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:

$\sqrt{\Delta}=\sqrt{297}=\sqrt{9*33}=\sqrt{9}*\sqrt{33}=3\sqrt{33}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-3\sqrt{33}}{2*18}=\frac{9-3\sqrt{33}}{36}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+3\sqrt{33}}{2*18}=\frac{9+3\sqrt{33}}{36}
$