3x^2+9=18x-3

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

3x^2+9=18x-3

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

3x^2-18x+12=0

a = 3; b = -18; c = +12;
Δ = b2-4ac
Δ = -182-4·3·12
Δ = 180
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:

$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-6\sqrt{5}}{2*3}=\frac{18-6\sqrt{5}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+6\sqrt{5}}{2*3}=\frac{18+6\sqrt{5}}{6}
$