Minimized

Let $f(x)=\dfrac{x^2}{14}+3x-5$ :

$f'(x)=\dfrac{x}{7}+3 \: \: f''(x)= \dfrac{1}{7}>0$

$f'(x)=0 \Rightarrow \dfrac{x}{7}+3=0 \Rightarrow x=-21$

As $f''(x)>0$ this is a minimum so the answer is:

$\boxed{x=-21}$

Just evaluate $\frac { -b }{ 2a } =-3\cdot 7=-21$ , which is the vertex of the parabola.

Similar solution as Ciprian Florea 's but in proper LaTex.

$\begin{aligned} \frac{x^2}{14} + 3x - 5 & = \frac{1}{14}(x^2 + 42x) - 5 \\ & = \frac{1}{14}(x^2 + 2(21)x + 21^2 - 21^2) - 5 \\ & = \frac{1}{14}(x+21)^2 - \frac{63}{2} - 5 \quad \quad \small \color{#3D99F6}{\text{Since }(x+21)^2 \ge 0} \\ & \ge - \frac{73}{2} \end{aligned}$

$\dfrac{x^2}{14} + 3x - 5$ is minimum when $x+21 = 0 \implies x = \boxed{-21}$