(2x+1)(5x-8)-(5x-8)(4x-7)=0

Solution for (2x+1)(5x-8)-(5x-8)(4x-7)=0 equation:

(2x+1)(5x-8)-(5x-8)(4x-7)=0

We multiply parentheses ..

(+10x^2-16x+5x-8)-(5x-8)(4x-7)=0

We get rid of parentheses

10x^2-16x+5x-(5x-8)(4x-7)-8=0

We multiply parentheses ..

10x^2-(+20x^2-35x-32x+56)-16x+5x-8=0

We add all the numbers together, and all the variables

10x^2-(+20x^2-35x-32x+56)-11x-8=0

We get rid of parentheses

10x^2-20x^2+35x+32x-11x-56-8=0

We add all the numbers together, and all the variables

-10x^2+56x-64=0

a = -10; b = 56; c = -64;
Δ = b2-4ac
Δ = 562-4·(-10)·(-64)
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(56)-24}{2*-10}=\frac{-80}{-20}
=+4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(56)+24}{2*-10}=\frac{-32}{-20}
=1+3/5
$