Question

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

Answer 1

what is the slope of y=-\frac{2}{3}x+5

Answer 2

To solve the quadratic equation \(x^2 + 32x + 248 = 0\), we can use the quadratic formula, which is given by:

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

In this equation, the coefficients are:
- \(a = 1\)
- \(b = 32\)
- \(c = 248\)

Now, we will substitute these values into the formula.

First, we calculate the discriminant:

\[
b^2 - 4ac = 32^2 - 4 \cdot 1 \cdot 248
\]

Calculating \(32^2\):

\[
32^2 = 1024
\]

Now calculating \(4 \cdot 1 \cdot 248\):

\[
4 \cdot 1 \cdot 248 = 992
\]

Now substitute back into the discriminant:

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

Now, plug the values back into the quadratic formula:

\[
x = \frac{-32 \pm \sqrt{32}}{2 \cdot 1}
\]

Calculating \(\sqrt{32}\):

\[
\sqrt{32} = \sqrt{16 \cdot 2} = 4\sqrt{2}
\]

Thus, substituting this back into the formula:

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

Now, simplifying this expression:

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

Therefore, the solutions to the quadratic equation \(x^2 + 32x + 248 = 0\) are:

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