First, from
$$\left(\frac{5}{29}\right)=\left(\frac{29}{5}\right)=\left(\frac{4}{5}\right)=\left(\frac{2}{5}\right)^{2}=1$$

we know that the given congruence equation must have a solution. Since $29 \equiv 5(\bmod 8)$, we get the solution from the above proof as
$$\begin{aligned} x & \equiv \pm\left(\frac{29-1}{2}\right)!\cdot 5^{\frac{29+3}{8}} \\ & = \pm(14!) \cdot 5^{4} \\ & \equiv \pm(3628800)(11)(12)(13)(14)(625) \\ & \equiv \pm(11)(12)(13)(14)(16) \\ & = \pm 384384 \equiv \pm 18(\bmod 29) \end{aligned}$$