Find all values of x such that

Ok lets see...

alright lets make these look better.

$\frac{x(x-4)}{(x-5)(x-4)}=\frac{4(x-5)}{(x-4)(x-5)}$

alright more simplifying:

$x(x-4)=4(x-5)$

$x^2-4x=4x-20$

$x^2-8x+20=0$

Oh yay! A quadratic equation! Now time to solve!

Ok so:

$x+y\:=\:-8,\:xy=20$

$x=-8-y$

Substitute this into xy = 20 and you will get

$\begin{pmatrix}x=-4+2i,\:&y=-4-2i\\ x=-4-2i,\:&y=-4+2i\end{pmatrix}$

so x is NOT 4±2i but it is -4\pm2i so yeah!

If anyone can check my work that would be great

The solutions are actually 3 \pm 2i.

We have:

x/x-5 = 4/x-4

Multiply by (x-5) and by (x-4) to both sides

x(x-4) = 4(x-5)

x^2 - 4x = 4x - 20

Put the left side on the right...

x^2 - 8x + 20 = 0

We find factors of 20 that multiply to 20, but add to -8 (we realize factors must be negative...)

Instead, we can just put this into our quadratic formula:

$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$

So we have:

8 +/- root(64 - 80) }/2

8 +/- root(-16) }/2

8 +/- 4i

4 +/- 2i

So the answer should be 4 +/- 2i, note that the 4 is positive...

(If this answer is incorrect please notify me as I would be happy to help) (But you did say this is incorrect, so I don't know what's going wrong here...)

$x(x-4) = 4(x-5)$

$x^2 - 4x = 4x - 20$

$x^2 - 8x + 20 = 0$

$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$

$x = {8 \pm \sqrt{-16} \over 2}$

$x = {8 \pm 4i \over 2}$

$\color{brown}\boxed{x = 4 \pm 2i}$

I think the problem had a typo. 

x^2 - 4x = 4x + 20

it gives imaginary, perhaps the owner of the question meant the right hand side to be 5/(x - 4)