Question

The product of three and a squared number is twice the sum of the number and four.

Answer 1

So basically you set this up as:

\[3*n^2 = 2(n+4)\]

Then solve for n:
\[3*n^2 = 2(n+4)\]\[3n^2 = 2n+8\]\[3n^2 - 2n -8 =0\]\[3n^2 - 6n + 4n - 8 = 0\]\[3n(n-2) + 4(n-2) = 0\]\[(3n+4)(n-2)=0\]

Then you know that:

\[3n+4 = 0\]So:
\[n = -4/3\]

AND

\[n-2=0\]So:
\[n=2\]

Therefore, n can equal either -4/3 or 2.