Question 306854
First complete the square to get to vertex form,
{{{2x^2-13x-7=0}}}
{{{2(x^2-(13/2)x)-7=0}}}
To complete the square, take (1/2) of the x-coefficient, square it, and then add it to both sides. 
Since this is going on inside the parenthese, make sure to multiply by 2 to add the right hand value.
{{{(1/2)(13/2)=13/4}}}
{{{(13/4)^2=169/16}}}
{{{2(x^2-(13/2)x+169/16)-7=2(169/16)}}}
{{{2(x-(13/4))^2-7=169/8)}}}
{{{2(x-(13/4))^2-56/8-169/8=0)}}}
{{{2(x-(13/4))^2-56/8-169/8=0)}}}
{{{2(x-(13/4))^2-225/8=0)}}}
Now the equation is in vertex form.
The vertex is ({{{13/4}}},{{{-225/8}}})
The y-intercept, from the original equation is, -7.
{{{ drawing( 300, 300, -3, 8, -30, 10, grid(1),circle(13/4,-225/8,.2),green(line(13/4,-50,13/4,50)),graph( 300, 300, -3, 8, -30, 10, 2x^2-13x-7)) }}} 
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Straight line on one side, what does that mean?
The axis of symmetry makes a mirror of the graph. 
So then 
{{{f(13/4+1)=f(13/4-1)}}}
{{{f(13/4+2)=f(13/4-2)}}}
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So then you only have to solve for one of those y coordinates and you'll have both.