The values of $y$ which will satisfy the equations $2x^{2}+6x+5y+1=0, 2x+y+3=0$ may be found by solving:
$\textbf{(A)}\ y^{2}+14y-7=0\qquad\textbf{(B)}\ y^{2}+8y+1=0\qquad\textbf{(C)}\ y^{2}+10y-7=0\qquad\\ \textbf{(D)}\ y^{2}+y-12=0\qquad \textbf{(E)}\ \text{None of these equations}$
If we solve the second equation for $x$ in terms of $y$, we find $x=-\dfrac{y+3}{2}$ which we can substitute to find:
\[2(-\dfrac{y+3}{2})^2+6(-\dfrac{y+3}{2})+5y+1=0\]
Multiplying by two and simplifying, we find:
\begin{align*} 2\cdot[2(-\dfrac{y+3}{2})^2+6(-\dfrac{y+3}{2})+5y+1]&=2\cdot 0\\ (y+3)^2 -6y-18+10y+2&=0\\ y^2+6y+9-6y-18+10y+2&=0\\ y^2+10y-7&=0 \end{align*}
Therefore the answer is $\boxed{\textbf{(C)}\ y^{2}+10y-7=0}$