There are two values of $k$ for which the equation $x^{2}+2 k x+7 k-10=0$ has two equal real roots (that is, has exactly one solution for $x$ ). The sum of these values of $k$ is
(A) 0
(B) -3
(C) 3
(D) -7
(E) 7
The equation $x^{2}+2 k x+7 k-10=0$ has two equal real roots precisely when the discriminant of this quadratic equation equals 0 .

The discriminant, $\Delta$, equals

$$
\Delta=(2 k)^{2}-4(1)(7 k-10)=4 k^{2}-28 k+40
$$

For the discriminant to equal 0 , we have $4 k^{2}-28 k+40=0$ or $k^{2}-7 k+10=0$ or $(k-2)(k-5)=0$. Thus, $k=2$ or $k=5$.

We check that each of these values gives an equation with the desired property.

When $k=2$, the equation is $x^{2}+4 x+4=0$ which is equivalent to $(x+2)^{2}=0$ and so only has one solution for $x$.

When $k=5$, the equation is $x^{2}+10 x+25=0$ which is equivalent to $(x+5)^{2}=0$ and so only has one solution for $x$.

The sum of these values of $k$ is $2+5=7$.

ANsWer: (E)