Answer :
To solve the quadratic equation [tex]\(3x^2 + 6x + 15 = 0\)[/tex], we will use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, the coefficients of the equation [tex]\(3x^2 + 6x + 15 = 0\)[/tex] are:
- [tex]\(a = 3\)[/tex]
- [tex]\(b = 6\)[/tex]
- [tex]\(c = 15\)[/tex]
1. Calculate the discriminant:
The discriminant [tex]\( \Delta \)[/tex] is given by [tex]\( \Delta = b^2 - 4ac \)[/tex]:
[tex]\[ \Delta = 6^2 - 4 \cdot 3 \cdot 15 = 36 - 180 = -144 \][/tex]
2. Determine the square root of the discriminant:
Since the discriminant is negative, the solutions will involve complex numbers. We take the square root of [tex]\(-144\)[/tex]:
[tex]\[ \sqrt{-144} = \sqrt{144} \cdot \sqrt{-1} = 12i \][/tex]
3. Apply the quadratic formula:
Substitute [tex]\(a = 3\)[/tex], [tex]\(b = 6\)[/tex], and [tex]\(\sqrt{\Delta} = 12i\)[/tex] into the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-6 \pm 12i}{2 \cdot 3} = \frac{-6 \pm 12i}{6} = \frac{-6}{6} \pm \frac{12i}{6} = -1 \pm 2i \][/tex]
So, the solutions to the quadratic equation [tex]\(3x^2 + 6x + 15 = 0\)[/tex] are:
[tex]\[ x = -1 + 2i \quad \text{and} \quad x = -1 - 2i \][/tex]
Thus, the correct answer is:
[tex]\[ -1 \pm 2i \][/tex]
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, the coefficients of the equation [tex]\(3x^2 + 6x + 15 = 0\)[/tex] are:
- [tex]\(a = 3\)[/tex]
- [tex]\(b = 6\)[/tex]
- [tex]\(c = 15\)[/tex]
1. Calculate the discriminant:
The discriminant [tex]\( \Delta \)[/tex] is given by [tex]\( \Delta = b^2 - 4ac \)[/tex]:
[tex]\[ \Delta = 6^2 - 4 \cdot 3 \cdot 15 = 36 - 180 = -144 \][/tex]
2. Determine the square root of the discriminant:
Since the discriminant is negative, the solutions will involve complex numbers. We take the square root of [tex]\(-144\)[/tex]:
[tex]\[ \sqrt{-144} = \sqrt{144} \cdot \sqrt{-1} = 12i \][/tex]
3. Apply the quadratic formula:
Substitute [tex]\(a = 3\)[/tex], [tex]\(b = 6\)[/tex], and [tex]\(\sqrt{\Delta} = 12i\)[/tex] into the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-6 \pm 12i}{2 \cdot 3} = \frac{-6 \pm 12i}{6} = \frac{-6}{6} \pm \frac{12i}{6} = -1 \pm 2i \][/tex]
So, the solutions to the quadratic equation [tex]\(3x^2 + 6x + 15 = 0\)[/tex] are:
[tex]\[ x = -1 + 2i \quad \text{and} \quad x = -1 - 2i \][/tex]
Thus, the correct answer is:
[tex]\[ -1 \pm 2i \][/tex]
