(2x+3)(4x-8)=(7x+1)(4x-8)

Solution for (2x+3)(4x-8)=(7x+1)(4x-8) equation:

(2x+3)(4x-8)=(7x+1)(4x-8)

We move all terms to the left:

(2x+3)(4x-8)-((7x+1)(4x-8))=0

We multiply parentheses ..

(+8x^2-16x+12x-24)-((7x+1)(4x-8))=0


We calculate terms in parentheses: -((7x+1)(4x-8)), so:

(7x+1)(4x-8)

We multiply parentheses ..

(+28x^2-56x+4x-8)

We get rid of parentheses

28x^2-56x+4x-8

We add all the numbers together, and all the variables

28x^2-52x-8

Back to the equation:

-(28x^2-52x-8)


We get rid of parentheses

8x^2-28x^2-16x+12x+52x-24+8=0

We add all the numbers together, and all the variables

-20x^2+48x-16=0

a = -20; b = 48; c = -16;
Δ = b2-4ac
Δ = 482-4·(-20)·(-16)
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-32}{2*-20}=\frac{-80}{-40}
=+2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+32}{2*-20}=\frac{-16}{-40}
=2/5
$