There I added the solution. Shouldn'that be added and subtracted but my prof has only added in the next step
Find the inverse of function $f(x)=x-5x^2$ when $x\ge 1$.
After completing the square somehow the signs don't match with the solution my professor did and it had me scratching my head.
$$f(x)=x-5x^2, x\ge 1$$
Due to $x \ge 1$, if $x_1, x_2$ are solutions of $y=x-5x^2$, the larger one is the valid preimage.
From $y=x-5x^2$, let's solve for $x$.
$$5x^2-x+y=0$$
From the quadratic formula, $$x=\frac{1+\sqrt{1-20y}}{10}$$
Hence, $$f^{-1}(x)=\frac{1+\sqrt{1-20x}}{10}$$