Question 1208849
Make the "change of variable" {{{y=x^2}}} and turn the equation into the quadratic equation
{{{y^2 + sqrt(2)y - 2 = 0}}}
Then you could use the quadratic formula that says that the solutions to
{{{ay^2+by+c=0}}} are given by {{{y = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
For the equation above {{{a=1}}}, {{{b=sqrt(2)}}}, and {{{c=-2}}}
You get {{{y = (-sqrt(2) + sqrt(2+8))/2=(-sqrt(2) + sqrt(10))/2=0.874032}}}(rounded), and
{{{y = (-sqrt(2) - sqrt(2+8))/2=(-sqrt(2) - sqrt(10))/2=-2.288246}}}(rounded)
At this point, if as I suspect you were not taught about complex and imaginary numbers, you can only list real solutions.
Then, {{{x^2=-2.288246}}} has no real solutions, and your solutions would be
{{{x=sqrt(0.874032)=0.93}}} and {{{x=-sqrt(0.874032)=-0.93}}}

NOTE:
If you are not fond of the quadratic formula, "completing the square" would be an option,
if you were taught to transform something like {{{y^2 + sqrt(2)y - 2 = 0}}}
into {{{y^2 + sqrt(2)y = 2 }}}, add {{{(sqrt(2)/2)^2}}} to both sides
to get {{{y^2 + sqrt(2)y +(sqrt(2)/2)^2= 2 +(sqrt(2)/2)^2}}},
rewrite it as {{{(y+sqrt(2)/2)^2=5/2}}} and {{{(y+sqrt(2)/2)^2=10/4}}} for the solutions
{{{y +sqrt(2)/2 = " " +- sqrt(10)/2}}} and {{{y =(-sqrt(2) +- sqrt(10))/2}}} 

They really quote that formula as {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} for {{{ax^2+bx+c=0}}},
but the name/letter used for the variable makes no difference.