$$n(t)=17t^2-20t+700$$ $$y=n(t)$$ $$y=17t^2-20t+700$$ 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?
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Basic approach is fine. Switching $y,t$ is somewhat confusing. Minor error towards end: $700-\frac{100}{17}=\frac{11800}{17}$ not $\frac{1180}{17}$. – almagest Jun 06 '16 at 13:16
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Edited the error. Thanks. And generally switching $y,t$ is just confusing or we shouldn't switch like that? – Siddharth Venu Jun 06 '16 at 13:19