How can I solve the quadratic equation 2x² + 15x + 1?

by

To solve the quadratic equation 2x² + 15x + 1 = 0, we can use the quadratic formula:

x = (-b ± √(b² – 4ac)) / 2a

In this equation, a, b, and c are the coefficients from the standard form of a quadratic equation, which is ax² + bx + c = 0. For our specific equation:

  • a = 2
  • b = 15
  • c = 1

Now, let’s plug these values into the quadratic formula:

  1. Calculate the value of b² – 4ac:

    2. 15² – 4(2)(1) = 225 – 8 = 217

  2. Now, substitute b and the result of the previous step into the formula:

    3. x = (−15 ± √217) / (2 * 2)

  3. Calculate the two potential solutions for x:

    4. x = (−15 + √217) / 4 and x = (−15 – √217) / 4

Now let’s evaluate these two expressions:

  • The first solution would be:
    x ≈ (−15 + 14.7) / 4 ≈ (−0.3) / 4 ≈ −0.075
  • The second solution would be:
    x ≈ (−15 – 14.7) / 4 ≈ (−29.7) / 4 ≈ −7.425

Therefore, the solutions for the quadratic equation 2x² + 15x + 1 = 0 are:

  • x ≈ −0.075
  • x ≈ −7.425

These values can be checked by substituting them back into the original equation to ensure they satisfy the equation.

Leave a Comment