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

1 answer

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

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

For the equation \(2x^2 - 5x - 3\), the coefficients are:

  • \(a = 2\)
  • \(b = -5\)
  • \(c = -3\)

Now, substitute these values into the quadratic formula:

  1. Calculate the discriminant:

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

  1. Now substitute back into the quadratic formula:

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

  1. Now we compute the two possible values for \(x\):
  • For the positive root:

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

  • For the negative root:

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

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

\[ x = 3 \quad \text{and} \quad x = -\frac{1}{2} \]