To solve the quadratic equation \(2x^2 - 5x - 3 = 0\), we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In this case, \(a = 2\), \(b = -5\), and \(c = -3\).
First, we calculate the discriminant (\(b^2 - 4ac\)):
\[ b^2 - 4ac = (-5)^2 - 4 \cdot 2 \cdot (-3) = 25 + 24 = 49 \]
Now we can find the solutions:
\[ x = \frac{-(-5) \pm \sqrt{49}}{2 \cdot 2} = \frac{5 \pm 7}{4} \]
Now we compute the two possible values for \(x\):
\[ x_1 = \frac{5 + 7}{4} = \frac{12}{4} = 3 \]
\[ x_2 = \frac{5 - 7}{4} = \frac{-2}{4} = -\frac{1}{2} \]
Thus, the solutions to the equation \(2x^2 - 5x - 3 = 0\) are:
\[ x = 3 \quad \text{and} \quad x = -\frac{1}{2} \]
You can enter these solutions where required. If you have any other questions or need further assistance, let me know!
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