Question

If possible write four solutions for the equation y=x^2+7

Answer 1

$$\begin{gathered} \mathrm{Hi}, \\ \mathrm{y}={\mathrm{x}}^2+7 \\ \mathrm{now}\mathrm{when}\mathrm{x}=0,1,2,3, \\ \mathrm{then}\mathrm{y}=0+7=7 \\ \mathrm{y}=1+7=8 \\ \mathrm{y}=2^2+7=11 \\ \mathrm{y}=3^2+7=16 \\ \mathrm{so}\mathrm{the}\mathrm{solution}\mathrm{sets}\mathrm{are}\left(0,7\right),\left(1,8\right),\left(2,11\right)\mathrm{and}\left(3,16\right) \end{gathered}$$