The quadratic $8x^2+12x-14$ has two real roots. What is the sum of the squares of these roots? Express your answer as a common fraction in lowest terms.
Let's call the two roots $r$ and $s$. We know that $r+s=-\frac{12}{8} = -\frac{3}{2}$. We also know that $rs=\frac{-14}{8} = -\frac{7}{4}$. We want to find $r^2 + s^2$. That's equal to $(r+s)^2 - 2rs$. So that's equal to $(-\frac{3}{2})^2 - 2(-\frac{7}{4})$. Which is equal to $\frac{9}{4} + \frac{7}{2}$. So the answer is $\frac{9}{4} + \frac{14}{4} = \frac{23}{4}$.

