If $\sin\alpha + \cos\alpha = 0.2$, find the numerical value of $\sin2\alpha$. 
If $\sin\alpha + \cos\alpha = 0.2$, find the numerical value of $\sin2\alpha$. How do I find a value for $\sin\alpha$ or $\cos\alpha$ so I can use a double angle formula?

I know how to solve a problem like
"If $\cos\alpha = \frac{\sqrt{3}}{2}$ , find $\sin2\alpha$" 
by using the 'double angle' formula: $\sin2\alpha = 2\sin\alpha\cos\alpha$ like this:
Start by computing $\sin\alpha$
$$\sin^2\alpha = 1 -\cos^2\alpha = 1-(\frac{\sqrt{3}}{2})^2 = \frac{1}{4}$$
so
$$\sin\alpha = \pm\frac{1}{2}$$
then it's just a simple matter of plugging $\sin\alpha = \pm\frac{1}{2}$ and $\cos\alpha=\frac{\sqrt{3}}{2}$ into $$\sin2\alpha = 2\sin\alpha\cos\alpha$$  to get $$\sin2\alpha = \pm\frac{\sqrt{3}}{2}$$
Where I can not make progress with the question 
"If $\sin\alpha + \cos\alpha = 0.2$, find the numerical value of $\sin2\alpha$".
Is how do I find a value for $\sin\alpha$ or $\cos\alpha$ so I can use a double angle formula?
What I have tried:
If $\sin\alpha+\cos\alpha = 0.2$
then $\sin\alpha=0.2-\cos\alpha$
and $\cos\alpha=0.2-\sin\alpha$. Should I start by  by computing $\sin\alpha$ using
$\sin^2\alpha = 1 -\cos^2\alpha = 1-(0.2-\cos\alpha)^2$?

$$\sin\alpha + \cos\alpha = 0.2$$
$$\sin\alpha + \sqrt {1-\sin^2\alpha}= 0.2$$
$$\sqrt{1-\sin^2\alpha}= 0.2-\sin\alpha$$
$$1-\sin^2\alpha=0.04-0.4\sin\alpha+\sin^2\alpha$$
$$2\sin^2\alpha-0.4\sin\alpha-0.96=0$$
$$\sin^2\alpha-0.2\sin\alpha-0.48=0$$
$$\sin\alpha=\frac{0.2\pm1.392...}{2}$$
$$2\sin\alpha=1,592...$$
$$2\sin\alpha=-1,192...$$
