The equation given is 
where a is some number.
We can solve for x by taking the square root of both sides.
Now let's think through what happens for various values of a.
TWO SOLUTIONS
If a is a positive number the above yields two solutions. Take for example:
There will be two solutions (one positive and one negative) as there are two numbers (here -7 and +7) that when multiplied by themselves give 49. That is,
and 
. The positive root is called the principal root and the negative root is called the secondary root. This will be the case anytime we take the root of a positive number.
ONE SOLUTION
If a = 0 there is only one solution. That is because
and 
. Zero is neither positive nor negative and it has only one root which is 0 itself. So in this case there is only one solution and it is 0.
NO (REAL) SOLUTIONS
If a is negative we would be taking the square root of a negative number. There is no (real) number that when multiplied by itself gives a negative number. Take for example
which gives us 
. The square root of -49 is not 7 because (7)(7)=49 which is positive. The square root of -49 is not -7 because (-7)(-7)=49 which is also positive. There is no real number that gives -49 when multiplied by itself. I say "real" numbers because there do exist imaginary/complex numbers but because of the way the questions was asked I imagine you may not know about these yet.
We can solve for x by taking the square root of both sides.
Now let's think through what happens for various values of a.
TWO SOLUTIONS
If a is a positive number the above yields two solutions. Take for example:
There will be two solutions (one positive and one negative) as there are two numbers (here -7 and +7) that when multiplied by themselves give 49. That is,
ONE SOLUTION
If a = 0 there is only one solution. That is because
NO (REAL) SOLUTIONS
If a is negative we would be taking the square root of a negative number. There is no (real) number that when multiplied by itself gives a negative number. Take for example