To solve the inequality 2x^2 + 3x + 1 ≥ 0, we can follow these steps:

Find the roots of the quadratic equation 2x^2 + 3x + 1 = 0 by factoring or using the quadratic formula.

The roots are x = -0.5 and x = -1.

Plot these roots on a number line to divide the number line into three intervals: x < -1, -1 ≤ x ≤ -0.5, and x > -0.5.

Test a value from each interval in the original inequality to determine where it is true.

For x < -1, use x = -2:
2(-2)^2 + 3(-2) + 1 = 8 - 6 + 1 = 3
3 > 0, so x < -1 is a solution.

For -1 ≤ x ≤ -0.5, use x = -1:
2(-1)^2 + 3(-1) + 1 = 2 - 3 + 1 = 0
0 ≥ 0, so -1 ≤ x ≤ -0.5 is a solution.

For x > -0.5, use x = 0:
2(0)^2 + 3(0) + 1 = 1
1 > 0, so x > -0.5 is a solution.

Therefore, the solution to the inequality 2x^2 + 3x + 1 ≥ 0 is x ≤ -1 or -0.5 ≤ x ≤ ∞.