Maximizing profit given cost and demand functions

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To maximize profit given the cost function C(x) = 13000 + 600x - 0.6x^2 + 0.004x^3 and the demand function p(x) = 1800 - 6x, the first step is to derive the revenue function R(x) = xp(x), resulting in R(x) = 1800x - 6x^2. The next step involves finding the marginal revenue by differentiating R(x) to get R'(x) = 1800 - 12x. Setting the marginal revenue equal to the marginal cost, derived from the cost function C(x), leads to the equation 1800 - 12x = 600 - 1.2x + 0.012x^2. The resulting quadratic equation can be solved for x to determine the production level that maximizes profit.
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Homework Statement



If
C(x) = 13000 + 600x − 0.6x62 + 0.004x3 is the cost function and
p(x) = 1800 − 6x is the demand function, find the production level that will maximize profit. (Hint: If the profit is maximized, then the marginal revenue equals the marginal cost.)

Homework Equations



R(x)=xp(x)

The Attempt at a Solution


Because of the hint, How I thought this question should be answered was to find the revenue function and take the derivative of that which is the marginal revenue and then set it equal to the derivative of the cost function which is marginal cost
So,
R(x)=x(1800-6x)
R(x)=1800x-6x^2
R'(x)=1800-12x

C'(x)=600-1.2x+.012x^2
1800-12x=600-1.2x+.012x^2
1200=10.8x+.012x^2
1200=.012x(900+x)
100,000=x(900+x)
100,000=900x+x^2

is this right? Where do I go from here, can I solve for x at this point? If so how?
 
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