(2x-3)(2x-3)=27

Solution for (2x-3)(2x-3)=27 equation:

(2x-3)(2x-3)=27

We move all terms to the left:

(2x-3)(2x-3)-(27)=0

We multiply parentheses ..

(+4x^2-6x-6x+9)-27=0

We get rid of parentheses

4x^2-6x-6x+9-27=0

We add all the numbers together, and all the variables

4x^2-12x-18=0

a = 4; b = -12; c = -18;
Δ = b2-4ac
Δ = -122-4·4·(-18)
Δ = 432
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:

$\sqrt{\Delta}=\sqrt{432}=\sqrt{144*3}=\sqrt{144}*\sqrt{3}=12\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-12\sqrt{3}}{2*4}=\frac{12-12\sqrt{3}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+12\sqrt{3}}{2*4}=\frac{12+12\sqrt{3}}{8}
$