(2x-1)(2x+1)-(x-3)(x+1)=18

Solution for (2x-1)(2x+1)-(x-3)(x+1)=18 equation:

(2x-1)(2x+1)-(x-3)(x+1)=18

We move all terms to the left:

(2x-1)(2x+1)-(x-3)(x+1)-(18)=0

We use the square of the difference formula

4x^2-(x-3)(x+1)-1-18=0

We multiply parentheses ..

4x^2-(+x^2+x-3x-3)-1-18=0

We add all the numbers together, and all the variables

4x^2-(+x^2+x-3x-3)-19=0

We get rid of parentheses

4x^2-x^2-x+3x+3-19=0

We add all the numbers together, and all the variables

3x^2+2x-16=0

a = 3; b = 2; c = -16;
Δ = b2-4ac
Δ = 22-4·3·(-16)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-14}{2*3}=\frac{-16}{6}
=-2+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+14}{2*3}=\frac{12}{6}
=2
$