Sorry, did not notice that the problem required to be solved using the "completing the squares" method.

Okay, here is the solution:

2x^2 - x - 6 = 0 x^2 - x/2 - 3 = 0 x^2 - x/2 = 3

Now we need to re write the left side of the equation in the form of x^2 + 2bx + b^2 So the equation becomes: x^2 + 2(-1/4)x + 1/16 - 1/16 = 3 (x^2 + 2(-1/4)x + 1/16) - 1/16 = 3 (x-1/4)^2 - 1/16 = 3 (x-1/4)^2 = 3 + 1/16 (x-1/4)^2 = 49/16

So there are two solutions: (x-1/4) = +√(49/16) (x-1/4) = -√(49/16)

In the first case (x-1/4) = +√(49/16) (x-1/4) = 7/4 x = 7/4 + 1/4 x = 8/4 x = 2

In the second case (x-1/4) = -√(49/16) (x-1/4) = -7/4 x = -7/4 + 1/4 x = -6/4 x = -1.5