Question 909625
{{{graph(300,300,-10,10,-10,10,(1/2)(x-2)^2+1,2x^2-6x-3)}}}
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Set the two functions equal to each other and solve for {{{x}}}.
{{{(1/2)(x-2)^2+1=2x^2-6x-3}}}
{{{(x-2)^2+2=4x^2-12x-6}}}
{{{x^2-4x+4+2=4x^2-12x-6}}}
{{{3x^2-8x-10=0}}}
{{{3(x^2-(8/3)x)-12=0}}}
{{{3(x^2-(8/3)x+16/9)-10=3(16/9)}}}
{{{3(x-4/3)^2=48/9+108/9}}}
{{{3(x-4/3)^2=156/9}}}
{{{(x-4/3)^2=52/9}}}
{{{x-4/3=0 +- sqrt(52)/3}}}
{{{x=4/3 +- (2sqrt(13))/3}}}
Now that you have the {{{x}}} coordinates of the intersection point, plug the value into either equation and solve for the {{{y}}} coordinates.
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Not sure I understand what C is.
Function {{{f(x)}}} does not cross the x-axis.
It's vertex is at (2,1) and it opens upwards. 
If you meant {{{g(x)}}} then,
{{{g(x)=0}}}
{{{2x^2-6x-3=0}}}
{{{2(x^2-3x)-3=0}}}
{{{2(x^2-3x+9/4)=2(9/4)+12/4}}}
{{{2(x-3/2)^2=30/4}}}
{{{(x-3/2)^2=15/4}}}
{{{x-3/2=0 +- sqrt(15)/2}}}
{{{x=3/2 +- sqrt(15)/2}}}