(2x-1)(2x+1)+(3x-1)(2x-3)=6x-1

Solution for (2x-1)(2x+1)+(3x-1)(2x-3)=6x-1 equation:

(2x-1)(2x+1)+(3x-1)(2x-3)=6x-1

We move all terms to the left:

(2x-1)(2x+1)+(3x-1)(2x-3)-(6x-1)=0

We use the square of the difference formula

4x^2+(3x-1)(2x-3)-(6x-1)-1=0

We get rid of parentheses

4x^2+(3x-1)(2x-3)-6x+1-1=0

We multiply parentheses ..

4x^2+(+6x^2-9x-2x+3)-6x+1-1=0

We add all the numbers together, and all the variables

4x^2+(+6x^2-9x-2x+3)-6x=0

We get rid of parentheses

4x^2+6x^2-9x-2x-6x+3=0

We add all the numbers together, and all the variables

10x^2-17x+3=0

a = 10; b = -17; c = +3;
Δ = b2-4ac
Δ = -172-4·10·3
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-13}{2*10}=\frac{4}{20}
=1/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+13}{2*10}=\frac{30}{20}
=1+1/2
$