(6x-3)(6x-3)=32

Solution for (6x-3)(6x-3)=32 equation:

(6x-3)(6x-3)=32

We move all terms to the left:

(6x-3)(6x-3)-(32)=0

We multiply parentheses ..

(+36x^2-18x-18x+9)-32=0

We get rid of parentheses

36x^2-18x-18x+9-32=0

We add all the numbers together, and all the variables

36x^2-36x-23=0

a = 36; b = -36; c = -23;
Δ = b2-4ac
Δ = -362-4·36·(-23)
Δ = 4608
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:

$\sqrt{\Delta}=\sqrt{4608}=\sqrt{2304*2}=\sqrt{2304}*\sqrt{2}=48\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-48\sqrt{2}}{2*36}=\frac{36-48\sqrt{2}}{72}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+48\sqrt{2}}{2*36}=\frac{36+48\sqrt{2}}{72}
$