(2x-3)(2x+1)=5

Solution for (2x-3)(2x+1)=5 equation:

(2x-3)(2x+1)=5

We move all terms to the left:

(2x-3)(2x+1)-(5)=0

We multiply parentheses ..

(+4x^2+2x-6x-3)-5=0

We get rid of parentheses

4x^2+2x-6x-3-5=0

We add all the numbers together, and all the variables

4x^2-4x-8=0

a = 4; b = -4; c = -8;
Δ = b2-4ac
Δ = -42-4·4·(-8)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-12}{2*4}=\frac{-8}{8}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+12}{2*4}=\frac{16}{8}
=2
$