3x^2+216=9x^2+18x

Solution for 3x^2+216=9x^2+18x equation:

3x^2+216=9x^2+18x

We move all terms to the left:

3x^2+216-(9x^2+18x)=0

We get rid of parentheses

3x^2-9x^2-18x+216=0

We add all the numbers together, and all the variables

-6x^2-18x+216=0

a = -6; b = -18; c = +216;
Δ = b2-4ac
Δ = -182-4·(-6)·216
Δ = 5508
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:

$\sqrt{\Delta}=\sqrt{5508}=\sqrt{324*17}=\sqrt{324}*\sqrt{17}=18\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-18\sqrt{17}}{2*-6}=\frac{18-18\sqrt{17}}{-12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+18\sqrt{17}}{2*-6}=\frac{18+18\sqrt{17}}{-12}
$