Finding the Standard Deviation of a Poisson Given A Relationship Between Probabilities

Suppose a certain random variable
$X$
follows a Poisson distribution with unknown standard deviation
$\sigma$
(for a Poisson distribution the variance equal the mean).
The following relationship between probabilities is known:
$P(X=2)-P(X=1)=3P(X=0)$
.
Using
$P(X=x)=\frac{e^{- \lambda} \lambda^x}{x!}$
we obtain
$\frac{e^{- \lambda} \lambda^2}{2!}-\frac{e^{- \lambda} \lambda^1}{1!}=3 \frac{e^{- \lambda} \lambda^0}{0!}$

$\frac{\lambda^2}{2}-\lambda=3$

$\lambda^2-2\lambda-6=0$

Hence
$\lambda = \frac{--2 \pm \sqrt{(-2)^2 - 4 \times 1 \times -6}}{2 \times 1} = \frac{2 \pm \sqrt{28}}{2} = 1 \pm \sqrt{7}$
.
The standard deviation is
$\sqrt{\lambda } = \sqrt{\frac{2 \pm \sqrt{28}}{2} = \sqrt{1 \pm \sqrt{7}}}$
and since
$1 - \sqrt{7} \lt 0$
(and so has no real square root) the standard deviation is
$\sigma = \sqrt{ \lambda} = \sqrt{1 + \sqrt{7}}$
.