ACT Math Practice Question: Modified quadratic equation

Try this ACT math practice question on a modified quadratic equation and how the resulting roots are related to the original quadratic equation.

The quadratic equation $ax^2-11x-14=0$ has two real solutions at $-\dfrac{2}{3}$ and $\dfrac{7}{5}$. What are the two solutions of the equation $a(x+1)^2-11(x+1)-14=0$?

1. $\quad -\dfrac{2}{3} \textrm{ and } \dfrac{7}{5}$
2. $\quad -\dfrac{5}{3} \textrm{ and } \dfrac{2}{5}$
3. $\quad -\dfrac{1}{3} \textrm{ and } \dfrac{12}{5}$
4. $\quad -\dfrac{2}{5} \textrm{ and } \dfrac{5}{3}$
5. $\quad -\dfrac{12}{5} \textrm{ and } \dfrac{1}{3}$