To find the solutions for the quadratic equation \(3x^2 + 8x + 4 = 0\), we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where:
- \(a = 3\)
- \(b = 8\)
- \(c = 4\)
First, we calculate the discriminant (\(b^2 - 4ac\)):
\[ b^2 - 4ac = 8^2 - 4 \cdot 3 \cdot 4 = 64 - 48 = 16 \]
Now we can apply the values to the quadratic formula:
\[ x = \frac{-8 \pm \sqrt{16}}{2 \cdot 3} = \frac{-8 \pm 4}{6} \]
This results in two solutions:
-
For the plus sign: \[ x_1 = \frac{-8 + 4}{6} = \frac{-4}{6} = -\frac{2}{3} \]
-
For the minus sign: \[ x_2 = \frac{-8 - 4}{6} = \frac{-12}{6} = -2 \]
Thus, the solutions of the quadratic equation \(3x^2 + 8x + 4 = 0\) are:
\[ x = -\frac{2}{3} \quad \text{and} \quad x = -2 \]