Question: Solve the quadratic equation below. If the solutions are not real, enter NA.
6x² − x − 2 = 0
The field below accepts a list of numbers or formulas separated by semicolons (e.g. 2; 4; 6 or
x + 1; x − 1). The order of the list does not matter.
To enter \sqrt{a}, type sqrt(a).
x =
1. **State the problem:** Solve the quadratic equation $$6x^2 - x - 2 = 0$$ for real values of $$x$$.
2. **Recall the quadratic formula:** For an equation $$ax^2 + bx + c = 0$$, the solutions are given by
$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$
where $$a = 6$$, $$b = -1$$, and $$c = -2$$.
3. **Calculate the discriminant:**
$$
\Delta = b^2 - 4ac = (-1)^2 - 4 \times 6 \times (-2) = 1 + 48 = 49
$$
Since $$\Delta > 0$$, there are two distinct real solutions.
4. **Apply the quadratic formula:**
$$
x = \frac{-(-1) \pm \sqrt{49}}{2 \times 6} = \frac{1 \pm 7}{12}
$$
5. **Find each solution:**
- For the plus sign:
$$
x = \frac{1 + 7}{12} = \frac{8}{12} = \frac{\cancel{4} \times 2}{\cancel{4} \times 3} = \frac{2}{3}
$$
- For the minus sign:
$$
x = \frac{1 - 7}{12} = \frac{-6}{12} = \frac{\cancel{-6} \times 1}{\cancel{6} \times 2} = -\frac{1}{2}
$$
6. **Final answer:** The solutions to the quadratic equation are $$x = \frac{2}{3}$$ and $$x = -\frac{1}{2}$$.