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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+3x^2-12x+x-4)-(2(x+1)(x-4))=0


We calculate terms in parentheses: -(2(x+1)(x-4)), so:

2(x+1)(x-4)

We multiply parentheses ..

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

We multiply parentheses

2x^2-8x+2x-8

We add all the numbers together, and all the variables

2x^2-6x-8

Back to the equation:

-(2x^2-6x-8)


We get rid of parentheses

3x^2-2x^2-12x+x+6x-4+8=0

We add all the numbers together, and all the variables

x^2-5x+4=0

a = 1; b = -5; c = +4;
Δ = b2-4ac
Δ = -52-4·1·4
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-3}{2*1}=\frac{2}{2}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+3}{2*1}=\frac{8}{2}
=4
$