(2x+3)(2x-5)-(x+3)(x-5)=

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

(2x+3)(2x-5)-(x+3)(x-5)=

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+4x^2-10x+6x-15)-(x+3)(x-5)=0

We get rid of parentheses

4x^2-10x+6x-(x+3)(x-5)-15=0

We multiply parentheses ..

4x^2-(+x^2-5x+3x-15)-10x+6x-15=0

We add all the numbers together, and all the variables

4x^2-(+x^2-5x+3x-15)-4x-15=0

We get rid of parentheses

4x^2-x^2+5x-3x-4x+15-15=0

We add all the numbers together, and all the variables

3x^2-2x=0

a = 3; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·3·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*3}=\frac{0}{6}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*3}=\frac{4}{6}
=2/3
$