The antiderivative of the acceleration $x″$ is the velocity

$x\prime \left(t\right)\=5 \frac{t^2}{2}\+3 t\+c$ 

The additive constant $c$ can be determined to be $c\=-7$ by solving the equation $x\prime \left(0\right)\=-7$ for $c$.

The antiderivative of the velocity $x\prime \left(t\right)\=\frac{5}{2} t^2\+3 t-7$ is the position

$x\left(t\right)\=\frac{5}{2} \frac{t^3}{3}\+3 \frac{t^2}{2}-7 t\+C$ 

Note the use of $C$ as the second additive constant introduced. This makes it easier to distinguish between the steps of the calculation. Finally, the constant $C$ can be determined to be $C\=8$ by solving the equation $x\left(0\right)\=8$ for $C$. Hence, the position function is

$x\left(t\right)\=\frac{5}{6}{t}^3\+\frac{3}{2}{t}^2-7 t\+8$