<PRE><B>Question 887314</B>
So first we can make a substitution.  Lets let s = x^2.  Then the equation becomes:&lt;br&gt;

s^2+2x-8 = 0&lt;br&gt;

Now we can factor the quadratic.&lt;br&gt;

s^2+2x-8 = 0 
(s-2)(s+4)=0&lt;br&gt;

Now we can plug in our value for s and set each factor equal to 0.&lt;br&gt;

(s-2)(s+4)=0
(x^2-2)(x^2+4)=0
x^2-2 = 0 and x^2+4 = 0
x^2 = 2 and x^2 = -4&lt;br&gt;

NOTE:Remember that the square root of a negative number is the same as the square root of the positive number times i.&lt;br&gt;

x = {{{sqrt(2)}}}, {{{-sqrt(2)}}} and x = 2i,-2i</PRE>