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

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

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

We multiply parentheses ..

(+6x^2+3x-10x-5)-(3x-5)(7x+3)=0

We get rid of parentheses

6x^2+3x-10x-(3x-5)(7x+3)-5=0

We multiply parentheses ..

6x^2-(+21x^2+9x-35x-15)+3x-10x-5=0

We add all the numbers together, and all the variables

6x^2-(+21x^2+9x-35x-15)-7x-5=0

We get rid of parentheses

6x^2-21x^2-9x+35x-7x+15-5=0

We add all the numbers together, and all the variables

-15x^2+19x+10=0

a = -15; b = 19; c = +10;
Δ = b2-4ac
Δ = 192-4·(-15)·10
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-31}{2*-15}=\frac{-50}{-30}
=1+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+31}{2*-15}=\frac{12}{-30}
=-2/5
$