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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+2x^2+10x-4x-20)-((2x-4)(2x-4))=0


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

(2x-4)(2x-4)

We multiply parentheses ..

(+4x^2-8x-8x+16)

We get rid of parentheses

4x^2-8x-8x+16

We add all the numbers together, and all the variables

4x^2-16x+16

Back to the equation:

-(4x^2-16x+16)


We get rid of parentheses

2x^2-4x^2+10x-4x+16x-20-16=0

We add all the numbers together, and all the variables

-2x^2+22x-36=0

a = -2; b = 22; c = -36;
Δ = b2-4ac
Δ = 222-4·(-2)·(-36)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-14}{2*-2}=\frac{-36}{-4}
=+9
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+14}{2*-2}=\frac{-8}{-4}
=+2
$