Help!

If the two roots of the quadratic $7x^2+3x+k$ are $\frac{-3\pm i\sqrt{299}}{14}$,

what is $k$?

$\begin{array}{|rcll|} \hline 7x^2+3x+k &=& 0 \quad | \quad : 7 \\\\ x^2+\frac37x+ \underbrace{\frac{k}{7}}_{=x_1x_2} &=& 0 \\\\ \dfrac{k}{7} &=& x_1x_2 \quad | \quad x_1 = \dfrac{-3+ i\sqrt{299}}{14} \quad x_2 = \dfrac{-3- i\sqrt{299}}{14} \\\\ \dfrac{k}{7} &=& \dfrac{\left(-3+ i\sqrt{299}\right) }{14}\cdot \dfrac{\left(-3- i\sqrt{299}\right) }{14} \\\\ \dfrac{k}{7} &=& \dfrac{(-3+ i\sqrt{299})(-3- i\sqrt{299}) }{14\cdot 14} \\\\ \dfrac{k}{7} &=& \dfrac{9- i^2\cdot 299 }{14\cdot 14} \quad | \quad i^2 = -1 \\\\ \dfrac{k}{7} &=& \dfrac{9- (-1)\cdot 299 }{14\cdot 14} \\\\ \dfrac{k}{7} &=& \dfrac{9+\cdot 299 }{14\cdot 14} \\\\ \dfrac{k}{7} &=& \dfrac{308}{14\cdot 14} \\\\ k &=& \dfrac{7\cdot 308}{14\cdot 14} \\\\ k &=& \dfrac{7\cdot 22}{14} \\\\ k &=& \dfrac{ 22}{2} \\\\ \mathbf{k} & \mathbf{=} & \mathbf{11} \\ \hline \end{array}$