Answer
$y=\pm\frac{1}{2}(x-4)$. See graph.
Work Step by Step
Step 1. Starting from the original equation and form squares with variables: $x^2+16=4(y^2+2x)$, $x^2-8x+16=4y^2$ or $(x-4)^2=4y^2$
Step 2. The above equation represents a degenerate conic: $4y^2=(x-4)^2$ or $y=\pm\frac{1}{2}(x-4)$ which gives a pair of lines.
Step 3. See graph.
