To find the roots of the quadratic equation [tex]\(2x^2 + 11x + 15 = 0\)[/tex], we will use the quadratic formula, which is given by:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
Here, the coefficients of the quadratic equation are:
- [tex]\(a = 2\)[/tex]
- [tex]\(b = 11\)[/tex]
- [tex]\(c = 15\)[/tex]
First, we need to calculate the discriminant, [tex]\( \Delta \)[/tex]:
[tex]\[
\Delta = b^2 - 4ac
\][/tex]
Substitute the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[
\Delta = 11^2 - 4 \cdot 2 \cdot 15 = 121 - 120 = 1
\][/tex]
Since the discriminant is a positive number, the quadratic equation has two real and distinct roots. Next, we use the quadratic formula to find the roots:
[tex]\[
x = \frac{-b \pm \sqrt{\Delta}}{2a}
\][/tex]
Substitute the values of [tex]\(b\)[/tex], [tex]\( \Delta \)[/tex], and [tex]\(a\)[/tex]:
[tex]\[
x = \frac{-(11) \pm \sqrt{1}}{2 \cdot 2} = \frac{-11 \pm 1}{4}
\][/tex]
This gives us two possible solutions:
1. Root 1:
[tex]\[
x = \frac{-11 + 1}{4} = \frac{-10}{4} = -2.5
\][/tex]
2. Root 2:
[tex]\[
x = \frac{-11 - 1}{4} = \frac{-12}{4} = -3
\][/tex]
Now, let's compare these roots with the available choices:
A. [tex]\(x = -6\)[/tex]
B. [tex]\(x = -3\)[/tex]
C. [tex]\(x = -5\)[/tex]
D. [tex]\(x = -\frac{5}{2}\)[/tex]
The roots we found are [tex]\(x = -2.5\)[/tex] and [tex]\(x = -3\)[/tex], which correspond to the choices:
B. [tex]\(x = -3\)[/tex]
D. [tex]\(x = -\frac{5}{2}\)[/tex] (which is the same as [tex]\(x = -2.5\)[/tex])
Therefore, the correct answers are:
[tex]\[
\boxed{x = -3 \text{ and } x = -\frac{5}{2}}
\][/tex]
