(2x+1)(2x-3)=4

Solution for (2x+1)(2x-3)=4 equation:

(2x+1)(2x-3)=4

We move all terms to the left:

(2x+1)(2x-3)-(4)=0

We multiply parentheses ..

(+4x^2-6x+2x-3)-4=0

We get rid of parentheses

4x^2-6x+2x-3-4=0

We add all the numbers together, and all the variables

4x^2-4x-7=0

a = 4; b = -4; c = -7;
Δ = b2-4ac
Δ = -42-4·4·(-7)
Δ = 128
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:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8\sqrt{2}}{2*4}=\frac{4-8\sqrt{2}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8\sqrt{2}}{2*4}=\frac{4+8\sqrt{2}}{8}
$