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

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

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

We move all terms to the left:

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

We use the square of the difference formula

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

We multiply parentheses ..

4x^2+(+6x^2+3x-8x-4)+1=0

We get rid of parentheses

4x^2+6x^2+3x-8x-4+1=0

We add all the numbers together, and all the variables

10x^2-5x-3=0

a = 10; b = -5; c = -3;
Δ = b2-4ac
Δ = -52-4·10·(-3)
Δ = 145
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{145}}{2*10}=\frac{5-\sqrt{145}}{20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{145}}{2*10}=\frac{5+\sqrt{145}}{20}
$