To find the roots of the quadratic equation [tex]\(2x^2 + 8x + 7 = 0\)[/tex], we can use the quadratic formula, which is given by:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
Here, the coefficients are [tex]\(a = 2\)[/tex], [tex]\(b = 8\)[/tex], and [tex]\(c = 7\)[/tex].
First, we need to calculate the discriminant, which is [tex]\(b^2 - 4ac\)[/tex]:
[tex]\[
\text{Discriminant} = b^2 - 4ac = 8^2 - 4 \cdot 2 \cdot 7 = 64 - 56 = 8
\][/tex]
So, the discriminant is [tex]\(8\)[/tex].
Next, we use the quadratic formula to find the roots:
[tex]\[
x = \frac{-b \pm \sqrt{\text{Discriminant}}}{2a} = \frac{-8 \pm \sqrt{8}}{2 \cdot 2} = \frac{-8 \pm \sqrt{8}}{4}
\][/tex]
Let's simplify [tex]\(\sqrt{8}\)[/tex]:
[tex]\[
\sqrt{8} = \sqrt{4 \cdot 2} = 2\sqrt{2}
\][/tex]
Now, we substitute [tex]\(\sqrt{8}\)[/tex] with [tex]\(2\sqrt{2}\)[/tex] in the quadratic formula:
[tex]\[
x = \frac{-8 \pm 2\sqrt{2}}{4} = \frac{-8}{4} \pm \frac{2\sqrt{2}}{4} = -2 \pm \frac{\sqrt{2}}{2}
\][/tex]
Upon further simplification:
[tex]\[
x = -2 \pm \frac{\sqrt{2}}{2}
\][/tex]
This gives us two roots:
[tex]\[
x_1 = -2 + \frac{\sqrt{2}}{2}
\][/tex]
And
[tex]\[
x_2 = -2 - \frac{\sqrt{2}}{2}
\][/tex]
From the given options, the correct answer matches the simplified form:
D. [tex]\( x = \frac{-2 \pm \sqrt{2}}{1} \)[/tex]
Hence, the correct answer is:
D. [tex]\(x = \frac{-2 \pm \sqrt{2}}{1}\)[/tex]
