(2x-5)(x-4)-(2x-5)(6x+1)=0

Solution for (2x-5)(x-4)-(2x-5)(6x+1)=0 equation:

(2x-5)(x-4)-(2x-5)(6x+1)=0

We multiply parentheses ..

(+2x^2-8x-5x+20)-(2x-5)(6x+1)=0

We get rid of parentheses

2x^2-8x-5x-(2x-5)(6x+1)+20=0

We multiply parentheses ..

2x^2-(+12x^2+2x-30x-5)-8x-5x+20=0

We add all the numbers together, and all the variables

2x^2-(+12x^2+2x-30x-5)-13x+20=0

We get rid of parentheses

2x^2-12x^2-2x+30x-13x+5+20=0

We add all the numbers together, and all the variables

-10x^2+15x+25=0

a = -10; b = 15; c = +25;
Δ = b2-4ac
Δ = 152-4·(-10)·25
Δ = 1225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1225}=35$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-35}{2*-10}=\frac{-50}{-20}
=2+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+35}{2*-10}=\frac{20}{-20}
=-1
$