Given 2x² + 4x + 3 = 0, we have a=2, b=4, c=3.

For a quadratic equation $ax^2 + bx + c = 0$, the sum of roots is $\alpha + \beta = -\frac{b}{a}$ and the product of roots is $\alpha\beta = \frac{c}{a}$.
Given $2x^2 + 4x + 3 = 0$, we have $a=2$, $b=4$, $c=3$.
Sum of roots:

$$\alpha + \beta = -\frac{4}{2} = -2$$

Product of roots:

$$\alpha\beta = \frac{3}{2}$$

Let the new roots be $r_1 = \alpha - 1$ and $r_2 = \beta - 1$.
Sum of new roots:

$$r_1 + r_2 = (\alpha - 1) + (\beta - 1) = \alpha + \beta - 2$$

Substitute the value of $\alpha + \beta$:

$$r_1 + r_2 = -2 - 2 = -4$$

Product of new roots:

$$r_1r_2 = (\alpha - 1)(\beta - 1) = \alpha\beta - \alpha - \beta + 1 = \alpha\beta - (\alpha + \beta) + 1$$

Substitute the values of $\alpha\beta$ and $\alpha + \beta$:

$$r_1r_2 = \frac{3}{2} - (-2) + 1 = \frac{3}{2} + 2 + 1 = \frac{3}{2} + 3 = \frac{3}{2} + \frac{6}{2} = \frac{9}{2}$$

A quadratic equation with roots $r_1$ and $r_2$ is given by $x^2 - (r_1 + r_2)x + r_1r_2 = 0$.
Substitute the sum and product of the new roots:

$$x^2 - (-4)x + \frac{9}{2} = 0$$ $$x^2 + 4x + \frac{9}{2} = 0$$

To eliminate the fraction, multiply the entire equation by 2:

$$2(x^2 + 4x + \frac{9}{2}) = 2(0)$$ $$2x^2 + 8x + 9 = 0$$

Comparing this with the given options, option (b) matches our result.