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Substitute $(5,0)$ and $(-3,0)$ into the equation:

we have:

$0=-2(5)^2+5b+c$ $0=-2(-3)^2-3b+c$

Combining the 2 equations:

$-2(5)^2+5b+c=-2(-3)^2-3b+c$ $-50+5b=-18-3b$ $8b=32$ $b=4$

Substitute $b=4$ into the first equation:

$0=-2(5)^2+5(4)+c$ $c=30$

Hence $b+c=4+30=\boxed{34}$

Slightly different from @Yan Yau Cheng :

First, for a parabola in the form $ax^2+bx+c$ , the axis of symmetry will always vertical. This means that given any two $x$ coordinates, the axis of symmetry will always be defined as $x=\frac{x_1+x_2}{2}$ where $x_1$ and $x_2$ are the $x$ -coordinates of two (not necessarily distinct) points sharing a common $y$ coordinate. Thus, we can average $5$ and $-3$ to find the axis of symmetry, which is thus $x=1$ . Now, we know that the formula for the axis of symmetry for a parabola is simply $\frac{-b}{2a}$ . Substituting, we get $1=\frac{b}{4} \Longrightarrow b=4$ .

Now, we can plug this in and solve for $c$ . Let's use $(5, 0)$ . Thus, $-2(25)+4(5)+c=0$ and $c=30$ . We can easily verify that this is also true for the point $(-3, 0)$ . Thus our desired answer is $30+4=\boxed{34}$ . Great problem Yan Yau! :D

Big story simple problem!LOLSimply put in those points, they would satisfy the equations!

Simply substitute values for x and y. Get a pair of linear equations.Solve them.Easy !!

Put y=0 to get a Quadratic equation in x it will be -2x^2+bx+c=0 Let the two roots be x1,x2 which are -3,5 Also x1+x2=b/2 x1*x2=(-2/c) Solve to get B=4 C=30