To solve the quadratic equation (6x^2 + x - 5 = 0), we can use the quadratic formula:
[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
]
In our equation, we have:
Now we will calculate the discriminant (b^2 - 4ac):
[
b^2 - 4ac = 1^2 - 4(6)(-5) = 1 + 120 = 121
]
Since the discriminant is positive, there will be two distinct real roots. Now we can plug the values into the quadratic formula:
[
x = \frac{-1 \pm \sqrt{121}}{2 \cdot 6}
]
Calculating (\sqrt{121}):
[
\sqrt{121} = 11
]
Now substituting back:
[
x = \frac{-1 \pm 11}{12}
]
This gives us two possible solutions:
- For the positive root:
[
x = \frac{-1 + 11}{12} = \frac{10}{12} = \frac{5}{6}
]
- For the negative root:
[
x = \frac{-1 - 11}{12} = \frac{-12}{12} = -1
]
Therefore, the solutions to the equation (6x^2 + x - 5 = 0) are:
[
x = \frac{5}{6} \quad \text{and} \quad x = -1
]
