Answer
$d = -7,d = 5$
Work Step by Step
Use either the square root method, the quadratic formula or factoring. $$
\begin{aligned}
d^2+2 d-35 & =0 \\
d^2+2 d+\left(\frac{2}{2}\right)^2 & =35+\left(\frac{2}{2}\right)^2 \\
d^2+2 d+1^2 & =35+1^2 \\
(d+1)^2 & =36 \\
d+1 & = \pm \sqrt{36} \\
d & =-1 \pm 6.
\end{aligned}
$$ This gives: $$
\begin{aligned}
d & =-1-6 \\
& =-7 \\
d & =-1+6 \\
& =5.
\end{aligned}
$$ Define the following version of the function and plot it. We see that the zeros are where they should be.
$$
T(d) =d^2+2 d-35.
$$
