SOLUTION: three-quarters times the square of a positive integer number is 3 less than fives times the integer. find the integer

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Question 39403: three-quarters times the square of a positive integer number is 3 less than fives times the integer. find the integer
Found 2 solutions by fractalier, Earlsdon:
Answer by fractalier(6550) About Me  (Show Source):
You can put this solution on YOUR website!
Let the number be x. Then
(3/4)x^2 = 5x - 3
Multiply by 4 to clear fractions...
3x^2 = 20x - 12
3x^2 - 20x + 12 = 0
(3x - 2)(x - 6) = 0
x = 2/3 or x = 6
but 2/3 is not an integer, so
x = 6.

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Translating the problem description into algebra:
%283%2F4%29n%5E2+=+5n-3 Simplify and solve for n. Mutliply through by 4 to clear the fraction.
3n%5E2+=+20n-12 Simplify.
3n%5E2-20n%2B12+=+0 Solve the quadratic equation by factoring.
%283n-2%29%28n-6%29+=+0 Apply the zero products principle.
3n-2+=+0 and/or n-6+=+0
If 3n-2+=+0 then 3n+=+2 and n+=+2%2F3 Discard this solution because you are looking for an integer.
If n-6+=+0 then n+=+6 This is the required integer.
Check:
%283%2F4%296%5E2+=+%283%2F4%2936 = 27
5%286%29-3+=+30-3 = 27