Finding out missing observation $a$ and $b$ The mean of $a,b,8,5,10$ is $6$ and the variance is $6.8$. find out value of $a$ and $b$


I tried something like this:

$$\bar{x}=6=\frac{a+b+8+5+10}{5}$$
$$7=a+b$$

Then i have put the value in the formula of variance

$$variance\ \sigma^2={\frac{{\Sigma x_i^2}}{n}-(\bar{x}^2)}$$
$$6.8={\frac{{a^2+b^2+189}}{5}-(36)}$$
$$25=a^2+b^2$$

I don't this my second equation is correct.If it correct then can anyone one guide me what do further?
You wrote the correct equation for $a^2+b^2$, but I get $$6.8={\frac{{a^2+b^2+189}}{n}-(36)}\\34=a^2+b^2+189-180\\25=a^2+b^2$$  Coupled with your correct value for $a+b$, you should be able to get home. 

Added: Now you have two equations in two unknowns.  You might be able to solve them by inspection.  Otherwise, you can write $b=7-a$ in substitute that into $25=a^2+b^2=a^2+(7-a)^2$, getting a quadratic in $a$.
Your second equation is wrong as well.

$$variance\ \sigma^2={\frac{{\Sigma (x_i - \bar{x})^2} }{n}}$$

That reduces to $$13 = (a-6)^2 + (b-6)^2$$ 

You can then substitute for either a or b from the first equation a + b = 7 to get a fairly simple quadtratic equation.