If I assume you meant supply is determined as `q_(s)=3p^2-4p` and demand is determined as `q_(d)=-p^2+24` then I would calculate that the equilibrium is where `q_(s)=q_(d)` and p>0. So I set up the equation `3p^2-4p=-p^2+24` I add `p^2` to both sides of the equation yielding `4p^2-4p=24` I subtract 24 from each side, yielding the quadratic equation `4p^2-4p-24=0` I factor the equation `(2p-6)(2p+4)=0` so that means `2p-6=0` or `2p+4=0` In the former case 2p=6 so p=3, and in the latter 2p=-4 so p=-2. The economic nature of the question requires p>0, so `p!=-2` and thus the equilibrium price is $3.00 each. Then to determine the equilibrium quantity, we solve either q expression for p=3.

`q_(s)=3(3)^2-4(3)=3(9)-12=27-12=15`

`q_(d)=-(3^2)+24=-9+24=15`

So the equilibrium quantity is 15 items.

The price and quantity for market equilibrium is 15 items at $3.00 each.

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