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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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


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

(3x-5)(x-3)

We multiply parentheses ..

(+3x^2-9x-5x+15)

We get rid of parentheses

3x^2-9x-5x+15

We add all the numbers together, and all the variables

3x^2-14x+15

Back to the equation:

-(3x^2-14x+15)


We get rid of parentheses

2x^2-3x^2+10x-3x+14x-15-15=0

We add all the numbers together, and all the variables

-1x^2+21x-30=0

a = -1; b = 21; c = -30;
Δ = b2-4ac
Δ = 212-4·(-1)·(-30)
Δ = 321
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-\sqrt{321}}{2*-1}=\frac{-21-\sqrt{321}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+\sqrt{321}}{2*-1}=\frac{-21+\sqrt{321}}{-2}
$