Question 320390
You're getting a quadratic equation because they intersect at two points. 
If they only intersected at one point, you would get a linear equation.
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 {{{ graph( 300, 300, -2,10, -10, 10, -2(x-3)^2 + 4, 2(x-2)^2 -1) }}} 
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 {{{-2(x-3)^2+4=2(x-2)^2 -1}}}
{{{-2(x^2-6x+9)+4=2(x^2-4x+4)-1}}}
{{{-2x^2+12x-18+4=2x^2-8x+8-1}}}
{{{4x^2-20x+21=0}}}
Factor
{{{(2x-7)(2x-3)=0}}}
Two solutions:
{{{2x-7=0}}}
{{{2x=7}}}
{{{x=7/2}}}
Then use either equation to find y.
{{{y=2(7/2-4/2)^2-1}}}
{{{y=2(3/2)^2-1}}}
{{{y=9/2-2/2}}}
{{{y=7/2}}}
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{{{2x-3=0}}}
{{{2x=3}}}
{{{x=3/2}}}
Again, find y,
{{{y=2(3/2-4/2)^2-1}}}
{{{y=2(-1/2)^2-1}}}
{{{y=-1/2-2/2}}}
{{{y=-1/2}}}
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{{{ drawing( 300, 300, -2,5, -2, 5, grid(1),
circle(3/2,-1/2,0.15),
circle(7/2,7/2,0.15),
graph( 300, 300, -2,5, -2, 5, -2(x-3)^2 + 4, 2(x-2)^2 -1)) }}} 
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(7/2,7/2) and (3/2,-1/2)