Question

Use the quadratic formula to solve the equation. $$-\frac{1}{2} x^{2}+6 x+13=0$$

Short answer

The solutions for the equation are \(x_1 = 2 - \sqrt{15}\; and \;x_2 = 2 + \sqrt{15}\)

Step-by-step solution

Identify the coefficients

From the equation \(-\frac{1}{2} x^{2}+6 x+13=0\), we can identify the coefficients as \(a = -\frac{1}{2}\), \(b = 6\), and \(c = 13\).

Substitute these coefficients into the quadratic formula

Replace \(a\), \(b\), and \(c\) in the formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(-\frac{1}{2}\), \(6\), and \(13\) respectively, we get \(x = \frac{-6 \pm \sqrt{6^2 - 4(-\frac{1}{2})*13}}{2*(-\frac{1}{2})}\).

Simplify the equation

Simplifying the above expression gives us two potential solutions for \(x\): \(x = \frac{-6 + \sqrt{36 + 26}}{-1}\) and \(x = \frac{-6 - \sqrt{36 + 26}}{-1}\)

Calculate the solutions

These are the calculated solutions: \(x_1 = 2 - \sqrt{15}\; and \;x_2 = 2 + \sqrt{15}\)