Question: Solve the quadratic equation below. If the solutions are not real, enter NA.
6x^2 - 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 =
Equation text: 6x^2 - x - 2 = 0, positioned top-center.
1. **State the problem:** Solve the quadratic equation $$6x^2 - x - 2 = 0$$ for $x$.
2. **Recall the quadratic formula:** For an equation $$ax^2 + bx + c = 0$$, the solutions are given by
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
3. **Identify coefficients:** Here, $a = 6$, $b = -1$, and $c = -2$.
4. **Calculate the discriminant:**
$$\Delta = b^2 - 4ac = (-1)^2 - 4 \times 6 \times (-2) = 1 + 48 = 49$$
5. **Since $\Delta > 0$, there are two real solutions.**
6. **Apply the quadratic formula:**
$$x = \frac{-(-1) \pm \sqrt{49}}{2 \times 6} = \frac{1 \pm 7}{12}$$
7. **Calculate 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:
$$x = \frac{1 - 7}{12} = \frac{-6}{12} = \frac{\cancel{-6} \times 1}{\cancel{6} \times 2} = -\frac{1}{2}$$
8. **Final answer:**
$$x = \frac{2}{3}; -\frac{1}{2}$$