Find values of the parameter a so that equation has equal roots. $x^2+2a\sqrt{a^2-3}x+4=0$
My final result was 2 and -0.5. Was it correct?

To get a pair of equal roots, we need to have  $2a\sqrt{a^2 - 3} = 4$.
$$\begin{align} 2a\sqrt{a^2-3}=4 &\iff \sqrt{a^4 - 3a^2}=  2 \\ \\ &\iff a^4 - 3a^2 = 4\\ \\ & \iff a^4 - 3a^2 - 4 = 0 \\ \\ & \iff (a^2 - 4)(a^2 +1) = 0\end{align} $$
$a^2 + 1 \geq 1$, so can never be zero. So we solve  $$a^2 - 4 = (a+2)(a-2)=0$$ $$ \iff a = -2 \text{ or } a = 2$$
