6x²+x-5=0

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:

• \(a = 6\)
• \(b = 1\)
• \(c = -5\)

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:

1. For the positive root:

\[ x = \frac{-1 + 11}{12} = \frac{10}{12} = \frac{5}{6} \]

2. 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:

\[ x = \frac{5}{6} \quad \text{and} \quad x = -1 \]