I thought the latex bragger died.

I thought this writing problem turned out above average. 

To find the vertex of the equation $y=-2x^2+bx+c$, we have to understand that the $x$ coordinate of the vertex can be found by taking the average of the x-intercepts and the $y$ coordinate can be found by letting $x=0$. Plugging it in leads to $y=c$, so we must find $c$ too.

So to find the $x$ intercepts, we have to let $y=0$ leading to a quadratic.

The sum of the roots of any quadratic is $-\frac{b}{a}$ from Vieta's formulas, and we can use this formula to help find $b$

Plugging in $b$ as $b$ and $a$ as $-2$, we get $\frac{-b}{-2}=\frac{b}{2}$
The sum of the roots can also be written as $(3+\sqrt5)+(3-\sqrt5)=6$

Since $6$ is the sum of the roots while $\frac{b}{2}$ can also be expressed as the sum of the roots, setting an equation will help us find $b$

$$\frac{b}{2}=6$$
$$b=12$$


So now our quadratic is $y=-2x^2+12x+c$



Now that found $b$ and $c$, the two vertexes can be found with ease.

The $x$ coordinate is found by taking the average of the sum of the roots, or $\frac{-b}{2a}$

Plugging in our values of $a$ and $b$ gives us:
$$\frac{-(-6)}{2(1)}$$
$$\frac{6}{2}$$
$$3$$

Since we now know that $x=3$, we can plug it in to the quadratic $y=-2x^2+12x+c$ to find the $y$ coordinate of the vertex.


Combining our two coordinates, our final vertex of the equation $y=-2x^2 + bx + c$ with roots $3+\sqrt5$ and $3-\sqrt5$ is $\boxed{\boxed{(3,10)}}$

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