Did I do the inverse of this quadratic function in the correct way?

Question

$$n(t)=17t^2-20t+700$$ $$y=n(t)$$ $$y=17t^2-20t+700$$ Switching $y$ and $t$, $$t=17y^2-20y+700$$ $$t=(17y^2-20y)+700 -(Eq. 1)$$ Completion of squares: $$\mathbf a=\sqrt{17}y$$ $$\mathbf b=???$$ $$\mathbf 2ab=-20y$$ $$\mathbf b=\frac{-20y}{2\sqrt{17}y}$$ $$\therefore b=\frac{-10}{\sqrt{17}}$$ $$\therefore(\sqrt{17}y-\frac{10}{\sqrt{17}})^2=17y^2-20y+\frac{100}{17}$$ Therefore add and subtract $b^2$ i.e., $\frac{100}{17}$ in $Eq. 1$: $$t=(17y^2-20y+\frac{100}{17})+700-\frac{100}{17}$$ $$\therefore t=(\sqrt{17}y-\frac{10}{\sqrt{17}})^2+\frac{11800}{17}$$ $$\therefore \pm\sqrt{t-\frac{11800}{17}}=\sqrt{17}y-\frac{10}{\sqrt{17}}$$ $$\therefore y=\frac{\frac{10}{\sqrt{17}}\pm\sqrt{t-\frac{11800}{17}}}{\sqrt{17}}$$ $$\therefore n^{-1}(t)=\frac{\frac{10}{\sqrt{17}}\pm\sqrt{t-\frac{11800}{17}}}{\sqrt{17}}$$ Is my approach and answer correct? And also, I feel completion of squares a little vulnerable to errors. How could have I solved the same without completion of squares?