(2x-3)(2x+3)=6

Solution for (2x-3)(2x+3)=6 equation:

(2x-3)(2x+3)=6

We move all terms to the left:

(2x-3)(2x+3)-(6)=0

We use the square of the difference formula

4x^2-9-6=0

We add all the numbers together, and all the variables

4x^2-15=0

a = 4; b = 0; c = -15;
Δ = b2-4ac
Δ = 02-4·4·(-15)
Δ = 240
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:

$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{15}}{2*4}=\frac{0-4\sqrt{15}}{8}
=-\frac{4\sqrt{15}}{8}
=-\frac{\sqrt{15}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{15}}{2*4}=\frac{0+4\sqrt{15}}{8}
=\frac{4\sqrt{15}}{8}
=\frac{\sqrt{15}}{2}
$