Question 327877
Since they both equal y, set them equal to each other.
{{{-x^2-2x+14=x^3-6x^2+12x-2}}}
{{{x^3-5x^2+14x-16=0}}}
{{{(x-2)(x^2-3x+8)=0}}}
The real solution is {{{x-2=0}}} or {{{x=2}}}.
Use either equation to solve for y.
{{{y=-(2)^2-2(2)+14}}}
{{{y=-4-4+14}}}
{{{y=6}}}
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{{{drawing(300,300,-10,10,-5,15,grid(1),circle(2,6,0.3),graph(300,300,-10,10,-5,15,-x^2-2x+14,x^3-6x^2+12x-2))}}}