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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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


We calculate terms in parentheses: -((x+5)(x-3)), so:

(x+5)(x-3)

We multiply parentheses ..

(+x^2-3x+5x-15)

We get rid of parentheses

x^2-3x+5x-15

We add all the numbers together, and all the variables

x^2+2x-15

Back to the equation:

-(x^2+2x-15)


We get rid of parentheses

2x^2-x^2+3x-10x-2x-15+15=0

We add all the numbers together, and all the variables

x^2-9x=0

a = 1; b = -9; c = 0;
Δ = b2-4ac
Δ = -92-4·1·0
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-9}{2*1}=\frac{0}{2}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+9}{2*1}=\frac{18}{2}
=9
$