How do I solve x + 2/3 = 2/x using the quadratic equation?

x + 2/3 = 2/x
x - 2/x + 2/3 = 0
3x(x - 2/x + 2/3) = 3x(0)
3x^2 - 6 + 2x = 0
3x^2 + 2x - 6 = 0
x = [-b ± √(b^2 - 4ac)]/(2a)

a = 3
b = 2
c = -6

x = [-2 ± √(4 + 72)]/6
x = [-2 ± √76]/6
x = [-2 ± √(2^2 * 19)]/6
x = [-2 ± 2√19]/6
x = [-1 ± √19]/3

∴ x = [-1 ± √19]/3

Question , as given , MUST be read as :-

x + (2/3) = 2/x

3 x ² + 2x - 6 = 0

x = [ - b ± √ ( b ² - 4 a c ) ] / 2 a

x = [ - 2 ± √ ( 4 + 72 ) ] / 2

x = [ - 2 ± √ ( 76 ) ] / 2

x = [ - 2 ± 2√19 ] / 2

x = - 1 ± √19

Multiply by x

x^2 + 2x/3 = 2

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

multiply by 3

3x^2 + 2x -6 =0

plug in A, B, C values

x = 1.1196329811802248 or -1.7862996478468913

tol, quadratic equation ba yan? sorry ah, la ako alam dyan eh :D