Question

Solve for x.
x^2+18=6x

Answer 1

That becomes:
\[x ^{2} - 6x + 18 = 0\]

Then solve for x using this equation:
\[x = \frac{b \pm \sqrt{b^2-4ac}}{2a}\]

So,
\[x = \frac{-6 \pm \sqrt{(-6)^2-4(1)(18)}}{2(1)}\]

But the value enclosed in the square root is negative. This makes it an imaginary number. You will end up with:

\[\frac{-6\pm6i}{2}\]