Question:

First, find $f(x)$ by substituting $x = 2t+1$ in the given equation. If $f(2x+1) = 4x^2 + 14x$, then let $2x+1 = t$, so $x = \frac{t-1}{2}$. Thus $f(t) = 4\left(\frac{t-1}{2}\right)^2 + 14\left(\frac{t-1}{2}\right) = (t-1)^2 + 7(t-1) = t^2 - 2t + 1 + 7t - 7 = t^2 + 5t - 6$. So $f(x) = x^2 + 5x - 6$. Setting $f(x) = 0$: $x^2 + 5x - 6 = 0$. Using Vieta's formulas, if roots are $r$ and $s$, then $r + s = -5$ and $rs = -6$. The sum of squares is $r^2 + s^2 = (r+s)^2 - 2rs = (-5)^2 - 2(-6) = 25 + 12 = 37$.