Question 22939
Yes there is. Quadratic equations are always solvable by using the quadratic formula: {{{x = (-b+-sqrt(b^2-4ac))/2a}}} 
For your equation, this would look like:

{{{x = (-4+-sqrt(4^2-4(1)(2)))/2(1)}}}
{{{x = (-4+-sqrt(16-8))/2}}}
{{{x = (-4+-sqrt(8))/2}}}
{{{x = (-4+-2sqrt(2))/2}}}
The roots are:
{{{x = -2+sqrt(2)}}}
{{{x = -2-sqrt(2)}}}