Just enter the equation:

(x-a)(x-b)

where a and b are the two roots. That, of course, is the quadratic equation with those roots

If a quadratic equation has two roots --let's call them x1 and x2--then the equation is proportional to

(x-x1)(x-x2)

In your example

x1 = Sqrt[10]

x2 = -Sqrt[10]

so the equaiton is

(x-Sqrt[10])(x+Sqrt[10])

you can expand this out to get the equivalent

x^2 - 10

My mistake, I wrote Sqrt[2] rather than Sqrt[10] ;-)

So I should have written

(x-Sqrt[10])(x+Sqrt[10])

I just edited my original post to correct this....

I do not believe that there is a way to enter the plus/minus character into Wolfram|Alpha. But when, in algebra, one uses that character it is just a shorthand for writing both cases.

I also note that one would hope that Wolfram|Alpha would give the answer to the statement

quadratic equation with roots 3 and 9

and give

27 - 12 x + x^2

as the result. But apparently the Wolfram|Alpha parser does note recognize this. I sent in a suggestion that it should address this case.