To solve the quadratic equation [tex]\(3u^2 + 5u + 2 = 0\)[/tex], we'll follow these steps:
1. Rewrite the equation in standard form:
The given equation is
[tex]\[
3u^2 = -5u - 2.
\][/tex]
To rewrite it in standard form [tex]\(ax^2 + bx + c = 0\)[/tex], we add [tex]\(5u\)[/tex] and [tex]\(2\)[/tex] to both sides:
[tex]\[
3u^2 + 5u + 2 = 0.
\][/tex]
2. Identify coefficients:
In the equation [tex]\(3u^2 + 5u + 2 = 0\)[/tex], the coefficients are:
- [tex]\(a = 3\)[/tex],
- [tex]\(b = 5\)[/tex],
- [tex]\(c = 2\)[/tex].
3. Apply the quadratic formula:
The quadratic formula is given by:
[tex]\[
u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.
\][/tex]
Plugging in the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[
u = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 3 \cdot 2}}{2 \cdot 3}.
\][/tex]
4. Simplify under the square root:
Calculate the discriminant:
[tex]\[
b^2 - 4ac = 5^2 - 4 \cdot 3 \cdot 2 = 25 - 24 = 1.
\][/tex]
Substitute back into the quadratic formula:
[tex]\[
u = \frac{-5 \pm \sqrt{1}}{6}.
\][/tex]
5. Simplify the solutions:
Taking the square root of 1:
[tex]\[
\sqrt{1} = 1.
\][/tex]
So, we have two possible solutions:
[tex]\[
u = \frac{-5 + 1}{6} \quad \text{and} \quad u = \frac{-5 - 1}{6}.
\][/tex]
6. Calculate the two solutions:
[tex]\[
u = \frac{-4}{6} = -\frac{2}{3},
\][/tex]
[tex]\[
u = \frac{-6}{6} = -1.
\][/tex]
Therefore, the solutions to the equation [tex]\(3u^2 + 5u + 2 = 0\)[/tex] are:
[tex]\[
u = -1 \quad \text{and} \quad u = -\frac{2}{3}.
\][/tex]
