Question 40 Solve the equation using the quadratic formula. \[ 3 x^{2}+x-5=0 \] (A) B \[ \left\{\frac{-1-\sqrt{61}}{2}, \frac{-1+\sqrt{61}}{2}\right\} \] (C) \[ \left\{\frac{1-\sqrt{61}}{6}, \frac{1+\sqrt{61}}{6}\right\} \] (D) \[ \left\{\frac{-1-\sqrt{61}}{6}, \frac{-1+\sqrt{61}}{6}\right\} \]

Step 1 ：Given the quadratic equation: \(3x^2 + x - 5 = 0\)

Step 2 ：Use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Step 3 ：Plug in the values: a = 3, b = 1, and c = -5

Step 4 ：Calculate the discriminant: \(b^2 - 4ac = 1^2 - 4(3)(-5) = 61\)

Step 5 ：Plug the discriminant into the quadratic formula: \(x = \frac{-1 \pm \sqrt{61}}{2(3)}\)

Step 6 ：Simplify the expression: \(x = \frac{-1 \pm \sqrt{61}}{6}\)

Step 7 ：\(\boxed{\left\{\frac{-1-\sqrt{61}}{6}, \frac{-1+\sqrt{61}}{6}\right\}}\)