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

Solution for (x-3)(x+5)=(2x+3)(x-4)=0 equation:

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

We move all terms to the left:

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

We multiply parentheses ..

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


We calculate terms in parentheses: -((2x+3)(x-4)), so:

(2x+3)(x-4)

We multiply parentheses ..

(+2x^2-8x+3x-12)

We get rid of parentheses

2x^2-8x+3x-12

We add all the numbers together, and all the variables

2x^2-5x-12

Back to the equation:

-(2x^2-5x-12)


We get rid of parentheses

x^2-2x^2+5x-3x+5x-15+12=0

We add all the numbers together, and all the variables

-1x^2+7x-3=0

a = -1; b = 7; c = -3;
Δ = b2-4ac
Δ = 72-4·(-1)·(-3)
Δ = 37
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{-(7)-\sqrt{37}}{2*-1}=\frac{-7-\sqrt{37}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{37}}{2*-1}=\frac{-7+\sqrt{37}}{-2}
$