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Let $p(x)=a{x}^2+bx+c.$ Then from the given information, $$\begin{array}{rl}4a-2b+c& =13,\\ a+b+c& =-2,\\ 9a+3b+c& =8.\end{array}$$Subtracting the first and second equations, and second and third equations, we get $$\begin{array}{rl}-3a+3b& =-15,\\ 8a+2b& =10.\end{array}$$Then $-a+b=-5$ and $4a+b=5.$ We can quickly solve, to find $a=2$ and $b=-3.$ Substituting into the equation $a+b+c=-2,$ we get $2-3+c=-2,$ so $c=-1.$ Therefore, $p(x)=\boxed{2x^2-3x-1}.$