We have that
\(-3x^2 + x + 3 = 9\).
First we should transform it so that the left hand side is equal to 0, and the \(x^2\) coefficient (that is, the number of \(x^2\)s you have) is equal to 1.
So first we will subtract 9, giving
\(-3x^2 + x -6 = 0\).
Now we will multiply by -1, giving
\(3x^2 - x + 6 = 0\),
and finally we will divide by 3 to make the \(x^2\) coefficient equal to 1, giving
\(x^2 - \frac{1}{3}x + 2 = 0\).
Now it is the form we need to complete the square. Notice that expanding \((x+p)^2\) gives
\(x^2 + 2px + p^2.\)
Now, we have \(\frac{1}{3} x\) in our equation so we want
\(2p = \frac{1}{3}
\implies p = \frac{1}{6}\).
If \(p = \frac{1}{6}\) then, from our expansion above, we have that
\((x+\frac{1}{6})^2 = x^2 + \frac{1}{3}x + \frac{1}{36}\),
which is almost our equation, but not quite. Instead of \(\frac{1}{36}\), we have 2. So we need to add \(\frac{71}{36}\), giving:
\((x+\frac{1}{6})^2 + \frac{71}{36} = x^2 + \frac{1}{3}x + 2 = 0\).
So, finally, subtracting the \(\frac{71}{36}\), we have
\((x+\frac{1}{6})^2 = -\frac{71}{36}\)
which is in the required form.
