0=-9x^2+18-3

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

0=-9x^2+18-3

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

9x^2-18+3=0

We add all the numbers together, and all the variables

9x^2-15=0

a = 9; b = 0; c = -15;
Δ = b2-4ac
Δ = 02-4·9·(-15)
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*9}=\frac{0-6\sqrt{15}}{18}
=-\frac{6\sqrt{15}}{18}
=-\frac{\sqrt{15}}{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{15}}{2*9}=\frac{0+6\sqrt{15}}{18}
=\frac{6\sqrt{15}}{18}
=\frac{\sqrt{15}}{3}
$