To solve the quadratic equation \(2x^2 + 5x + 3 = 0\) using the quadratic formula, we first identify the coefficients \(a\), \(b\), and \(c\):
- \(a = 2\)
- \(b = 5\)
- \(c = 3\)
The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Now let's calculate the discriminant \(b^2 - 4ac\):
\[ b^2 - 4ac = 5^2 - 4(2)(3) = 25 - 24 = 1 \]
Now we can substitute \(b\) and the calculated discriminant back into the quadratic formula:
\[ x = \frac{-5 \pm \sqrt{1}}{2(2)} \]
Since \(\sqrt{1} = 1\), we have:
\[ x = \frac{-5 \pm 1}{4} \]
This gives us two potential solutions for \(x\):
- \(x = \frac{-5 + 1}{4} = \frac{-4}{4} = -1\)
- \(x = \frac{-5 - 1}{4} = \frac{-6}{4} = -\frac{3}{2} = -1.5\)
Thus, the solutions rounded to the tenths are:
\[ x = -1.0 \quad \text{and} \quad x = -1.5 \]
So the final answers are:
\[ x = -1.0 \quad \text{and} \quad x = -1.5 \]