The two pieces are continuous for every
![c \in \mathbb{R}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/5ppe8aoin9q272gdu6xw915ao089fprtzz.png)
So, our only concern is to make sure that the pieces "glue" continuously at x=7.
To ensure this, we evaluate both pieces at x=7, and impose that the two values are equal.
The first piece evaluates to
![f_1(7)=7^2-c = 49-c](https://img.qammunity.org/2020/formulas/mathematics/middle-school/1piqtwynf534iunvchrdrajco5sntw34tl.png)
The second piece evaluates to
![f_2(7)=7c+1](https://img.qammunity.org/2020/formulas/mathematics/middle-school/6bdxjsvcan6lkgbac8676n814yj2eq5xbh.png)
So, we want
![7c+1=49-c](https://img.qammunity.org/2020/formulas/mathematics/middle-school/wdklgdnm9uwke5c9czvug65lzf8zkx2rex.png)
We move all terms involving c to the left hand side, and all the numerical constants on the right hand side:
![8c=48 \iff c=6](https://img.qammunity.org/2020/formulas/mathematics/middle-school/2xmlhgj8b7t16armve0l934pz8gwip8asg.png)