To determine the horizontal intercepts, we'll set $$y=0$$ and solve for $$x\text{:}$$

\begin{align*} 0\amp=-2(x+5)^2+12\\ \end{align*}

We cannot immediately factor this equation, but we can use the square root property to solve it. To do so, we'll first need to subtract $$12$$ from each side of the equation and then divide each side by $$-2\text{:}$$

\begin{align*} -12\amp=-2(x+5)^2\\ 6\amp=(x+5)^2\\ \end{align*}

Now applying the square root property,

\begin{align*} \pm\sqrt{6}\amp=x+5\\ x\amp=-\sqrt{6}-5,\ x=\sqrt{6}-5\\ x\amp\approx -7.449,\ x\approx -2.551 \end{align*}

The $$x$$-intercepts occur at $$(-\sqrt{6}-5,0)$$ and $$(\sqrt{6}-5,0)\text{,}$$ or at approximately $$(-7.449,0)$$ and $$(-2.551,0)\text{.}$$

To determine the $$y$$-intercept, we'll replace $$x$$ with $$0\text{:}$$

\begin{align*} y\amp=-2(\substitute{0}+5)^2+12\\ \amp=-2(25)+12\\ \amp=-38 \end{align*}

The $$y$$-intercept occurs at $$(0,-38)\text{.}$$

To determine the vertex using the formula $$x=-\frac{b}{2a}\text{,}$$ we'll first need to expand this equation so that $$a$$ and $$b$$ can be identified:

\begin{align*} y\amp=-2(x+5)^2+12\\ y\amp=-2(x^2+10x+25)+12\\ y\amp=-2x^2-20x-50+12\\ y\amp=-2x^2-20x-38 \end{align*}

Now we can use $$a=-2$$ and $$b=-20$$ to determine the $$x$$-value of the vertex:

\begin{align*} x\amp=-\frac{-20}{2(-2)}\\ \amp=-5 \end{align*}

Lastly, we'll use this value to find the $$y$$-value. We can use either the original equation or the expanded one; we'll use the original one here:

\begin{align*} y\amp=-2(\substitute{-5}+5)^2+12\\ \amp=-2(0)+12\\ \amp=12 \end{align*}

Therefore the vertex is the point $$(-5,12)\text{.}$$

in-context