Question 6533
Let a=b<sup>2</sup>. Then the equation becomes a<sup>2</sup>+a-20=0. Factoring leaves {{{(a+5)(a-4)=0}}} So a=-5 or a=4. So we have {{{b^2=-5}}} and {{{b^2=4}}} So the real solutions are b=2 and b=-2. (Note:The imaginary solutions are {{{b=i*root(2,5)}}} and {{{b=-i*root(2,5)}}}).