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

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

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

We move all terms to the left:

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

We use the square of the difference formula

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

We multiply parentheses ..

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


We calculate terms in parentheses: -((+3x^2+3x-4x-4)), so:

(+3x^2+3x-4x-4)

We get rid of parentheses

3x^2+3x-4x-4

We add all the numbers together, and all the variables

3x^2-1x-4

Back to the equation:

-(3x^2-1x-4)


We get rid of parentheses

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

We add all the numbers together, and all the variables

-2x^2+x=0

a = -2; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-2)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*-2}=\frac{-2}{-4}
=1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-2}=\frac{0}{-4}
=0
$