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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+X^2-5X-4X+20)-((X-4)(2X+3))=0


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

(X-4)(2X+3)

We multiply parentheses ..

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

We get rid of parentheses

2X^2+3X-8X-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-4X+5X+20+12=0

We add all the numbers together, and all the variables

-1X^2-4X+32=0

a = -1; b = -4; c = +32;
Δ = b2-4ac
Δ = -42-4·(-1)·32
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-12}{2*-1}=\frac{-8}{-2}
=+4
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+12}{2*-1}=\frac{16}{-2}
=-8
$