Algebra.Com's Answer #51909 by bucky(2189)![\"\"](\"/images/stars/stars-13.gif\")  
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document.write( "Substitute 9500 for C(x):
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document.write( "Subtract 9500 from both sides and transpose to get:
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document.write( "This will factor into:
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document.write( "and you set each of the two factors equal to zero to find that either x = 105 or x = -90 are
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document.write( "the solutions to this equation. And since it makes no sense to have a minus number of
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document.write( "units manufactured, the correct answer is x = 105 units.
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document.write( "But a more general solution exists. It is the quadratic formula and it is based on a method
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document.write( "called completing the squares. You can just know that it exists so you can look it up if you
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document.write( "need to use it, or you can (as most people that work math generally do) just memorize it.
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document.write( "This formula says that if you have a quadratic equation of the form:
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document.write( "the answers for x are given by:
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document.write( "By comparing with the equation we have you
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document.write( "can see that a = 1, b = -15, and c = -9450.
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document.write( "Substitute these numbers into the equation to solve for x and you get:
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document.write( "which simplifies to:
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document.write( "So the two answers for x are:
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document.write( "and
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document.write( "These are the same answers that we got by factoring, so we probably haven't made a mistake.
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document.write( "Notice that the quadratic formula is a general method of use. It will work on quadratic
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document.write( "equations when, as in this case, it is hard to find the factors or the numbers are so
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document.write( "big that it is difficult to do a graph with enough accuracy. It is well worth memorizing the
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document.write( "formula if you have lots of difficult quadratics to work on.
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document.write( "Hope this discussion helps to fill in the gaps left by the book you are using.
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