Question 41170
b^4+2b^2-24=0

let b^2 be t; then
b^4+2b^2-24=t^2+2t-24=0  (factor t^2+2t-24)

(t-4)(t+6)=0  
t-4=0 
t=4    and

t+6=0
t=-6  Since t=b^2

t=b^2 =4 so, b=2 and b=-2
t=b^2=-6 since square of a real number cannot be negative this equation has no solutions.

So, the solutions of this equation are 2 and -2.