Question 1144640
Where they intersection, {{{-x^2+3x+8=2x+6}}}

{{{-x^2+x+2=0}}}

{{{x^2-x-2=0}}}

{{{(x+1)(x-2)=0}}}

At x=-1 and x=2.


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{{{y=2(-1)+6=-2+6=4}}}
point (-1,4)
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and
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{{{y=2*2+6=10}}}
point (2,10)
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{{{graph(400,400,-6,12,-6,12,-x^2+3x+8,2x+6)}}}