Let the above quadratic have roots $x = p$ and $q$ such that by Vieta's Formulae:
$p + q = 2 - a$ (i);
$pq = -2a^2 + 5a - 3$ (ii);
$|p| = 2|q|$ (iii).
Squaring (i) yields $p^2 + 2pq + q^2 = 4 - 4a + a^2$, and substitution of (ii) into this expression gives us:
$p^2 + q^2 + 2(-2a^2 + 5a - 3) = 4 - 4a + a^2$;
or $4q^2 + q^2 - 4a^2 + 10a - 6 = 4 - 4a + a^2$;
or $5q^2 = 5a^2 - 14a + 10$;
or $|q| = \sqrt{(5a^2 -14a + 10)/5}$ (iii).
From (i), let $p = 2 - a - q = 2 - a \pm \sqrt{(5a^2 -14a + 10)/5}$ and substitute this expression into $|p| = 2|q|$ to yield:
$|2 - a \pm \sqrt{(5a^2 -14a + 10)/5}| = 2\sqrt{(5a^2 -14a + 10)/5}$;
or $2 - a \pm \sqrt{(5a^2 -14a + 10)/5} = \pm 2\sqrt{(5a^2 -14a + 10)/5}$;
or $2 - a = \pm \sqrt{(5a^2 -14a + 10)/5}$ OR $\pm 3\sqrt{(5a^2 -14a + 10)/5}$;
or $(2 - a)^2 = (5a^2 -14a + 10)/5$ OR $9 \cdot (5a^2 -14a + 10)/5$.
For the first case above, we obtain:
$(2 - a)^2 = (5a^2 -14a + 10)/5$;
or $5(4 - 4a + a^2) = 5a^2 -14a + 10$;
or $10 = 6a$;
or a = 5/3
and the second case above yields:
$(2 - a)^2 = 9 \cdot (5a^2 -14a + 10)/5$;
or $5(4 - 4a + a^2) = 45a^2 -26a + 90$;
or $0 = 40a^2 - 106a + 70$;
or $0 = 20a^2 - 53a + 35$;
or $0 = (4a - 5)(5a - 7)$;
or a = 5/4, 7/5
Thus, the complete solution is $\fbox{a = 5/4, 7/5, 5/3}$.