Question

1/6x^2 +6=0

Answer 1

\[\frac{1}{6}x^2+6=0\] like that?

Answer 2

not possible, at least not with real numbers, because \(x^2\geq 0\) always and so \(\frac{1}{6}x^2+6\geq 6\) for all \(x\)

Answer 3

yeah

Answer 4

you sure it isn't
\[\frac{1}{6}x^2-6=0\]?

Answer 5

yes im sure

Answer 6

16x^2-81=0