Finding all real solutions of $x-8\sqrt{x}+7=0$ Finding all real solutions of $x-8\sqrt{x}+7=0$.
Man, I tried subtituting $x=y^2$ but IDK things got complicated. What is the best way to figure this out? Thanks!

Let $y=\sqrt{x}$, therefore $y^2=x$
$y^2-8y+7=0$ therefore $(y-7)(y-1)=0$ hence $y=1,7$.
Therefore $x=1,49$. These are indeed the only solutions.
