1) Solve `2x^{2}-5x-3=0`and enter solutions below.

Question

2) Push the "Graph It" Button to see a graph of `y=2x^{2}-5x-3`
Enter solution 1 and 2

Answer 1

To solve the quadratic equation \(2x^{2}-5x-3=0\), we can use the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

where \(a = 2\), \(b = -5\), and \(c = -3\).

1. First, we calculate the discriminant (\(b^2 - 4ac\)):

\[ b^2 - 4ac = (-5)^2 - 4(2)(-3) = 25 + 24 = 49 \]

2. Now we can use the quadratic formula:

\[ x = \frac{-(-5) \pm \sqrt{49}}{2(2)} = \frac{5 \pm 7}{4} \]

3. We find the solutions:

• For the positive case:

\[ x_1 = \frac{5 + 7}{4} = \frac{12}{4} = 3 \]

• For the negative case:

\[ x_2 = \frac{5 - 7}{4} = \frac{-2}{4} = -\frac{1}{2} \]

Thus, the solutions to the equation \(2x^{2}-5x-3=0\) are:

1. \(x_1 = 3\)
2. \(x_2 = -\frac{1}{2}\)

You can enter these solutions as:

1. \(3\)
2. \(-\frac{1}{2}\)