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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

(2x+5)(x-4)-((x-4)(-1x+5))=0

We multiply parentheses ..

(+2x^2-8x+5x-20)-((x-4)(-1x+5))=0


We calculate terms in parentheses: -((x-4)(-1x+5)), so:

(x-4)(-1x+5)

We multiply parentheses ..

(-1x^2+5x+4x-20)

We get rid of parentheses

-1x^2+5x+4x-20

We add all the numbers together, and all the variables

-1x^2+9x-20

Back to the equation:

-(-1x^2+9x-20)


We get rid of parentheses

2x^2+1x^2-8x+5x-9x-20+20=0

We add all the numbers together, and all the variables

3x^2-12x=0

a = 3; b = -12; c = 0;
Δ = b2-4ac
Δ = -122-4·3·0
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{-(-12)-12}{2*3}=\frac{0}{6}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+12}{2*3}=\frac{24}{6}
=4
$