3. The equation is transformed into the form

$$
t^{\alpha}=\alpha(t+1), \quad \alpha=\lg 2 \in(0,1), \quad t=x^{2}-3>0
$$

where:

1) the left side of the equation is a concave up function (since $\alpha \in(0 ; 1)$), defined for $t \geqslant 0$
2) the right side of the equation is a linear function with a positive slope
3) at $t=0$, the value of the left side is less than the value of the right side; at $t=1$, the value of the left side is, on the contrary, greater than the value of the right side, since

$$
1^{\alpha}=\lg 10>\lg 4=\alpha \cdot(1+1)
$$

Therefore, the graphs intersect at two points (one between 0 and 1, the other to the right of 1), each of which generates two roots of the original equation.
Answer: 4. Answer to option: $1-2: 4$.

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