JAMB 1993 · UME · Q20

Bring everything to one side: $y^2 - 3y - 18 > 0$.

Factorise: $(y - 6)(y + 3) > 0$, so the roots are $y = 6$ and $y = -3$.

The graph of a positive $y^2$ quadratic is a U shape, so it is above zero outside the roots: $y < -3$ or $y > 6$, option B.