3x^2-18x^2+9=0

Solution for 3x^2-18x^2+9=0 equation:

3x^2-18x^2+9=0

We add all the numbers together, and all the variables

-15x^2+9=0

a = -15; b = 0; c = +9;
Δ = b2-4ac
Δ = 02-4·(-15)·9
Δ = 540
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:

$\sqrt{\Delta}=\sqrt{540}=\sqrt{36*15}=\sqrt{36}*\sqrt{15}=6\sqrt{15}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{15}}{2*-15}=\frac{0-6\sqrt{15}}{-30}
=-\frac{6\sqrt{15}}{-30}
=-\frac{\sqrt{15}}{-5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{15}}{2*-15}=\frac{0+6\sqrt{15}}{-30}
=\frac{6\sqrt{15}}{-30}
=\frac{\sqrt{15}}{-5}
$