I keep getting the wrong answer and I dont know what step im missing.

3x^{2} + 2x - 3 = 0

Since this quadratic equation is set equal to 0, we know we are looking to solve for the zeros/roots of the function. Note that these solutions are the x-intercepts (i.e., they are the points when y=0 on the graph of the function).

The easiest way to solve for the zeros of the function would be to factor the quadratic equation and then use the zero product property. This equation, however, cannot be factored. With this, we have the option solving by either completing the square or using the quadratic formula. Since a≠1, the latter would be the simpler option.

Quadratic formula: x = (-b ± √(b^{2} - 4ac))/2a

Given that a = 3 , b = 2 , and c = -3 , then

x = (-2 ± √(2^{2} - 4(3)(-3)))/(2)(3)

= (-2 ± √(4 + 36))/6

= (-2 ± √(40))/6

= (-2 ± √(4)·√(10))/6

= (-2 ± 2√(10))/6

= 2(-1 ± √10)/6

= (-1 ± √10)/3

Thus, the two solutions to this quadratic function are

x = (-1 + √10)/3 and x = (-1 - √10)/3