3(9x^2-36x+36)+9=27

Solution for 3(9x^2-36x+36)+9=27 equation:

3(9x^2-36x+36)+9=27

We move all terms to the left:

3(9x^2-36x+36)+9-(27)=0

We add all the numbers together, and all the variables

3(9x^2-36x+36)-18=0

We multiply parentheses

27x^2-108x+108-18=0

We add all the numbers together, and all the variables

27x^2-108x+90=0

a = 27; b = -108; c = +90;
Δ = b2-4ac
Δ = -1082-4·27·90
Δ = 1944
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:

$\sqrt{\Delta}=\sqrt{1944}=\sqrt{324*6}=\sqrt{324}*\sqrt{6}=18\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-108)-18\sqrt{6}}{2*27}=\frac{108-18\sqrt{6}}{54}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-108)+18\sqrt{6}}{2*27}=\frac{108+18\sqrt{6}}{54}
$