The real numbers $x, y$ and $z$ satisfy the three equations

$$
\begin{aligned}
x+y & =7 \\
x z & =-180 \\
(x+y+z)^{2} & =4
\end{aligned}
$$

If $S$ is the sum of the two possible values of $y$, then $-S$ equals
(A) 56
(B) 14
(C) 36
(D) 34
(E) 42
Since $x+y=7$, then $x+y+z=7+z$.

Thus, the equation $(x+y+z)^{2}=4$ becomes $(7+z)^{2}=4$.

Since the square of $7+z$ equals 4 , then $7+z=2$ or $7+z=-2$.

If $7+z=2$, then $z=-5$.

In this case, since $x z=-180$, we get $x=\frac{-180}{-5}=36$ which gives $y=7-x=-29$.

If $7+z=-2$, then $z=-9$.

In this case, since $x z=-180$, we get $x=\frac{-180}{-9}=20$ which gives $y=7-x=-13$.

We can check by direct substitution that $(x, y, z)=(36,-29,-5)$ and $(x, y, z)=(20,-13,-9)$ are both solutions to the original system of equations.

Since $S$ is the sum of the possible values of $y$, we get $S=(-29)+(-13)=-42$ and so $-S=42$.