Questions: Find the real solutions, if any, using the quadratic formula. 5x^2 - 13x + 6 = 0 Select the correct choice below and fill in any answer boxes in your choice. A. The solution set is (Simplify your answer. Use a comma to separate answers as needed.) B. The solution is not a real number.

Solution Steps

To find the real solutions of the quadratic equation \(5x^2 - 13x + 6 = 0\), we can use the quadratic formula, which is given by:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

where \(a\), \(b\), and \(c\) are the coefficients of the quadratic equation \(ax^2 + bx + c = 0\). In this case, \(a = 5\), \(b = -13\), and \(c = 6\). We will calculate the discriminant \(b^2 - 4ac\) to determine if the solutions are real. If the discriminant is non-negative, we will compute the solutions using the quadratic formula.

The given quadratic equation is \(5x^2 - 13x + 6 = 0\). The coefficients are:

• \(a = 5\)
• \(b = -13\)
• \(c = 6\)

The discriminant \(\Delta\) of a quadratic equation \(ax^2 + bx + c = 0\) is given by: \[ \Delta = b^2 - 4ac \] Substituting the values of \(a\), \(b\), and \(c\): \[ \Delta = (-13)^2 - 4 \times 5 \times 6 = 169 - 120 = 49 \]

Since the discriminant \(\Delta = 49\) is positive, the quadratic equation has two distinct real roots.

The quadratic formula is: \[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \] Substituting the values of \(b\), \(\Delta\), and \(a\): \[ x = \frac{-(-13) \pm \sqrt{49}}{2 \times 5} = \frac{13 \pm 7}{10} \]

Calculate the two possible values for \(x\):

1. \(x_1 = \frac{13 + 7}{10} = \frac{20}{10} = 2\)
2. \(x_2 = \frac{13 - 7}{10} = \frac{6}{10} = \frac{3}{5}\)

Final Answer