hWhat are the solutions to `x^{2}+32x+248\ =\ 0`?

Answer 1

To solve the quadratic equation \(x^{2} + 32x + 248 = 0\), we can use the quadratic formula:

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

where \(a = 1\), \(b = 32\), and \(c = 248\).

First, we need to calculate the discriminant (\(b^2 - 4ac\)):

\[ b^2 = 32^2 = 1024 \] \[ 4ac = 4 \times 1 \times 248 = 992 \] \[ b^2 - 4ac = 1024 - 992 = 32 \]

Now, we can apply the values in the quadratic formula:

\[ x = \frac{-32 \pm \sqrt{32}}{2 \times 1} \] \[ x = \frac{-32 \pm 4\sqrt{2}}{2} \] \[ x = \frac{-32}{2} \pm \frac{4\sqrt{2}}{2} \] \[ x = -16 \pm 2\sqrt{2} \]

Thus, the solutions to the equation \(x^{2} + 32x + 248 = 0\) are:

\[ x = -16 + 2\sqrt{2} \quad \text{and} \quad x = -16 - 2\sqrt{2} \]