Here's the question you clicked on:
leidyr
Solve p-3√(p) =10
now jst squaring both sides
u get quadratic in "p"
\[3\sqrt{p}=p-10\] \[3\sqrt{p}=p-10\] squre both sides \[9p=p^2-20p+100\] \[p^2-29p+100=0\] \[(p-25)(p-4)=0\] \[p=25,4\] p=4 4-6=10? false, discard p=4 p=25 25-15=10? true, p=25 is the solution