To find the roots of the equation [tex]\(2x - 5 = -3x^2\)[/tex], we need to rearrange it into a standard form of the quadratic equation, which is [tex]\(ax^2 + bx + c = 0\)[/tex].
Step-by-step solution:
1. Start with the given equation:
[tex]\[
2x - 5 = -3x^2
\][/tex]
2. Move all terms to one side of the equation to set it equal to 0:
[tex]\[
3x^2 + 2x - 5 = 0
\][/tex]
3. The resulting quadratic equation is:
[tex]\[
3x^2 + 2x - 5 = 0
\][/tex]
4. Next, we solve this quadratic equation for [tex]\(x\)[/tex]. The roots of a quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex] can be found using the quadratic formula:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
For the equation [tex]\(3x^2 + 2x - 5 = 0\)[/tex]:
- [tex]\(a = 3\)[/tex]
- [tex]\(b = 2\)[/tex]
- [tex]\(c = -5\)[/tex]
5. Substituting these values into the quadratic formula, we get two solutions for [tex]\(x\)[/tex]:
[tex]\[
x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 3 \cdot (-5)}}{2 \cdot 3}
\][/tex]
[tex]\[
x = \frac{-2 \pm \sqrt{4 + 60}}{6}
\][/tex]
[tex]\[
x = \frac{-2 \pm \sqrt{64}}{6}
\][/tex]
[tex]\[
x = \frac{-2 \pm 8}{6}
\][/tex]
6. Solving these two potential solutions, we get:
[tex]\[
x = \frac{-2 + 8}{6} = \frac{6}{6} = 1
\][/tex]
and
[tex]\[
x = \frac{-2 - 8}{6} = \frac{-10}{6} = -\frac{5}{3}
\][/tex]
The two roots of the quadratic equation [tex]\(3x^2 + 2x - 5 = 0\)[/tex] are:
[tex]\[
x = 1 \quad \text{and} \quad x = -\frac{5}{3}
\][/tex]
Therefore, the correct answers from the given options are:
- B. [tex]\(x = 1\)[/tex]
- D. [tex]\(x = -\frac{5}{3}\)[/tex]
So, the roots of the equation [tex]\(2x - 5 = -3x^2\)[/tex] are:
[tex]\[
\boxed{B \quad and \quad D}
\][/tex]
