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(x^2)/(x-5) - 1 = 25/(x-5)
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\[\frac{x^2}{x-5}-1=\frac{25}{x-5}\\ \frac{x^2}{x-5}-\frac{x-5}{x-5}=\frac{25}{x-5}\\ \frac{x^2-(x-5)}{x-5}=\frac{25}{x-5}\\ \frac{x^2-x+5}{x-5}=\frac{25}{x-5} \] Note that \(x=5\) can't be a solution. So if \(x\not=5\), you have \[x^2-x+5=25\\ x^2-x-20=0\]
Can you solve that any further?
Can you? It's not my problem to do. If you want a hint, it can be factored.
I didn't mean you like will you do it for me. I was just asking if it was possible.
If I factor the first one, do I have to make it equal 0 instead of 25?
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Nvm XD
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