(1-5x)(2x-3)+(2x-3)(x+7)=0

Solution for (1-5x)(2x-3)+(2x-3)(x+7)=0 equation:

(1-5x)(2x-3)+(2x-3)(x+7)=0

We add all the numbers together, and all the variables

(-5x+1)(2x-3)+(2x-3)(x+7)=0

We multiply parentheses ..

(-10x^2+15x+2x-3)+(2x-3)(x+7)=0

We get rid of parentheses

-10x^2+15x+2x+(2x-3)(x+7)-3=0

We multiply parentheses ..

-10x^2+(+2x^2+14x-3x-21)+15x+2x-3=0

We add all the numbers together, and all the variables

-10x^2+(+2x^2+14x-3x-21)+17x-3=0

We get rid of parentheses

-10x^2+2x^2+14x-3x+17x-21-3=0

We add all the numbers together, and all the variables

-8x^2+28x-24=0

a = -8; b = 28; c = -24;
Δ = b2-4ac
Δ = 282-4·(-8)·(-24)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-4}{2*-8}=\frac{-32}{-16}
=+2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+4}{2*-8}=\frac{-24}{-16}
=1+1/2
$