Question 182222
To find the intersection of the two curves, set them equal to eacgh other. 
{{{2x^2-5x-12=0.5x^2-3x+4}}}
{{{1.5x^2-2x-16=0}}}
{{{3x^2-4x-32=0}}}
You can factor this quadratic,
{{{(3x+8)(x-4)=0}}}
First solution:
3x-8=0
x=-8/3
Second solution:
x-4=0
x=4
Use either equation to find corresponding y values. 
{{{y(-8/3)=2(-8/3)^2-5(-8/3)-12=2(64/9)+40/3-12}}}
{{{y(-8/3)=(128/9)+(120/9)-(108/9)=(128+120-108)/9=140/9}}}
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{{{y(4)=2(4)^2-5(4)-12=2(16)-20-12=32-32=0}}}
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Intersection points : (-8/3,140/9) and (4,0)
{{{ graph( 300, 300, -8, 8, -2, 18, 2x^2-5x-12, 0.5x^2-3x+4) }}}