(3x-3)(3x-3)=26

Solution for (3x-3)(3x-3)=26 equation:

(3x-3)(3x-3)=26

We move all terms to the left:

(3x-3)(3x-3)-(26)=0

We multiply parentheses ..

(+9x^2-9x-9x+9)-26=0

We get rid of parentheses

9x^2-9x-9x+9-26=0

We add all the numbers together, and all the variables

9x^2-18x-17=0

a = 9; b = -18; c = -17;
Δ = b2-4ac
Δ = -182-4·9·(-17)
Δ = 936
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:

$\sqrt{\Delta}=\sqrt{936}=\sqrt{36*26}=\sqrt{36}*\sqrt{26}=6\sqrt{26}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-6\sqrt{26}}{2*9}=\frac{18-6\sqrt{26}}{18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+6\sqrt{26}}{2*9}=\frac{18+6\sqrt{26}}{18}
$