2y2 - 3y + 1 = 0

To solve the equation 2y^2 - 3y + 1 = 0, we can use the quadratic formula:

y = (-b ± √(b^2 - 4ac)) / (2a).

Here, a = 2, b = -3, and c = 1.

Substituting these values into the formula:

y = (-(-3) ± √((-3)^2 - 4 * 2 * 1)) / (2 * 2) = (3 ± √(9 - 8)) / 4 = (3 ± √1) / 4 = (3 ± 1) / 4.

Therefore, we have two solutions:

1. y = (3 + 1) / 4 = 4 / 4 = 1.
2. y = (3 - 1) / 4 = 2 / 4 = 1/2.

So, the solutions to the equation 2y^2 - 3y + 1 = 0 are y = 1 and y = 1/2.