Question 677673
{{{ F(x)=2x^2+8x+3 }}}
This is a parabola with a minimum and not a maximum
I can tell because the squared term is positive. If I changed it
to {{{ F(x)=-2x^2+8x+3 }}} then it would have a maximum.
The vertex is the (x,y) point at the minimum ( or maximum )
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The x-coordinate of the vertex occurs halfway 
between the roots,  which are
the two solutions to {{{ 2x^2+8x+3 = 0 }}}   
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To get a deeper understanding of this look at the 
quadratic formula which gives these two solutions
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{{{ x[1] = (-b + sqrt( b^2 - 4*a*c )) / (2*a) }}} 
{{{ x[2] = (-b - sqrt( b^2 - 4*a*c )) / (2*a) }}} 
Now I'll separate the terms in each solution:
{{{ x[1] = -b/(2a) + ( sqrt( b^2 - 4*a*c) ) / (2*a) }}} 
{{{ x[2] = -b/(2a) - ( sqrt( b^2 - 4*a*c) ) / (2*a) }}} 
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Now what you have is:
{{{ x[1] }}} = ( x coordinate of the vertex ) + ( the positive correction )
{{{ x[2] }}} = ( x coordinate of the vertex ) - ( the negative correction )
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So, you can see that {{{ -b/(2a) }}} is the x-coordinate of the vertex
when the form of the equation is {{{ f(x) = ax^2 + b*x + c }}}
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I can apply this to your problem
{{{ F(x)=2x^2+8x+3 }}}
{{{ a = 2 }}}
{{{ b = 8 }}}
{{{ c = 3 }}}
{{{ -b/(2a) = -8/(2*2) }}}       
{{{ -b/(2a) = -2 }}}
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Now you know the vertex is at ( -2, y ). To find y, plug {{{ x = -2 }}}
into the equation
{{{ y =2*(-2)^2+8*(-2)+3 }}}
{{{ y = 2*4 - 16 + 3 }}}
{{{ y = 8 - 16 + 3 }}}
{{{ y = -5 }}}
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The vertex is at ( -2, -5 )
Here's a plot of the equation:
{{{ graph( 400, 400, -7, 7, -7, 7, 2x^2 + 8x + 3 ) }}}
The vertex is where is should be