Answer
The solutions are $-\sqrt3, -\sqrt2, \sqrt2, \text{ and } \sqrt3$.
Work Step by Step
The given equation can be written as:
$(a^2)^2-5a^2+6=0$
Let $u=a^2$.
Rewrite the equation above using $u$ to obtain:
$u^2-5u+6=0$
Factor the trinomial to obtain:
$(u-3)(u-2)=0$
Factor each factor with a variable to zero then solve each equation to obtain:
\begin{array}{ccc}
&u-3=0 &\text{or} &u-2=0
\\&u=3 &\text{or} &u=2
\end{array}
Since $u=a^2$, then:
\begin{array}{ccc}
&u=3 &\text{or} &u=-\frac{5}{2}
\\&a^2=3 &\text{or} &a^2=2
\\&(u=\pm\sqrt3 &\text{or} &u=\pm \sqrt2
\end{array}
Therefore, the solutions are $-\sqrt3, -\sqrt2, \sqrt2, \text{ and } \sqrt3$.