Binomial Distribution: Finding the number of trials given probability and successess

If $X$ is the number of successful trials, then assuming independence of trials $X$ has a Binomial$(n,p)$ distribution where $n$ is the number of trials. Then the probability of at least two successes is \begin{align} P[X \geq 2] &= 1 - P[X = 0] - P[X = 1] \\ &= 1 - \binom{n}{0}p^0q^n - \binom{n}{1}p^1q^{n-1} \\ &= 1 - q^n - npq^{n-1} \\ &= 1 - (n+1)\left(\frac{1}{2}\right)^n. \end{align} Setting this to be greater than or equal to $0.9$ gives $$ 0.1 \geq (n+1)\left(\frac{1}{2}\right)^n. $$ It happens that this occurs first at $n=7$.