To solve the quadratic equation [tex]\(2x^2 + 16x - 8 = 0\)[/tex], we need to use the quadratic formula:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
Here, [tex]\(a = 2\)[/tex], [tex]\(b = 16\)[/tex], and [tex]\(c = -8\)[/tex].
First, let's calculate the discriminant:
[tex]\[
\Delta = b^2 - 4ac
\][/tex]
Substituting the values, we get:
[tex]\[
\Delta = 16^2 - 4 \times 2 \times (-8)
\][/tex]
[tex]\[
\Delta = 256 + 64
\][/tex]
[tex]\[
\Delta = 320
\][/tex]
Now, we find the square root of the discriminant:
[tex]\[
\sqrt{\Delta} = \sqrt{320} = \sqrt{64 \times 5} = 8\sqrt{5}
\][/tex]
Next, apply the quadratic formula:
[tex]\[
x = \frac{-16 \pm 8\sqrt{5}}{2 \times 2}
\][/tex]
Simplify the expression:
[tex]\[
x = \frac{-16 \pm 8\sqrt{5}}{4}
\][/tex]
[tex]\[
x = \frac{-16}{4} \pm \frac{8\sqrt{5}}{4}
\][/tex]
[tex]\[
x = -4 \pm 2\sqrt{5}
\][/tex]
Thus, the solutions to the quadratic equation [tex]\(2x^2 + 16x - 8 = 0\)[/tex] are:
[tex]\[
x = -4 + 2\sqrt{5} \quad \text{and} \quad x = -4 - 2\sqrt{5}
\][/tex]
The correct answer is:
A. [tex]\(x = -4 \pm 2\sqrt{5}\)[/tex]
