Yes, is it possible. Since the product of two negative numbers is a positive numbers, for any real number x you have
![x \cdot x = x^2,\text{ but also }\ (-x)(-x) = x^2](https://tex.z-dn.net/?f=%20x%20%5Ccdot%20x%20%3D%20x%5E2%2C%5Ctext%7B%20but%20also%20%7D%5C%20%28-x%29%28-x%29%20%3D%20x%5E2%20)
So, a number and its opposite give the same result when squared. For example, both 3 and -3 give 9 when squared.
So, when a variable is squared into an equation, you will have two different solutions. For example, in the equation
![x^2 = 25](https://tex.z-dn.net/?f=%20x%5E2%20%3D%2025%20)
You are looking for a number that gives 25 when squared. We know that 5 gives 25 when squared, but because of everything we said at the beginning, -5 also gives 25 when squared. So, the two solutions are 5 and -5.
About the last question, I assume that by "an equation with a squared variable" you mean a quadratic equation, i.e. something like
![ax^2+bx+c=0](https://tex.z-dn.net/?f=%20ax%5E2%2Bbx%2Bc%3D0%20)
In this case, there can be two different solutions, one double solution, or no solution at all. In fact, the solving formula involves a square root, namely
![\sqrt{b^2-4ac}](https://tex.z-dn.net/?f=%20%5Csqrt%7Bb%5E2-4ac%7D%20)
and a root doesn't always exist.