What are the roots of y = x² + 3x + 10?

To find the roots of the quadratic equation y = x² + 3x + 10, we need to use the quadratic formula:

x = (-b ± √(b² – 4ac)) / 2a

In this equation, a is the coefficient of , b is the coefficient of x, and c is the constant term. Here, we have:

  • a = 1
  • b = 3
  • c = 10

First, we calculate the discriminant (b² – 4ac):

Discriminant = 3² – 4(1)(10) = 9 – 40 = -31

Since the discriminant is negative, this indicates that there are no real roots for the equation. Instead, we have two complex roots. We can proceed to find these complex roots:

x = (-3 ± √(-31)) / 2(1)

We know that √(-1) can be replaced with the imaginary unit i, so:

√(-31) = √31 * i

Substituting this back into our formula gives:

x = (-3 ± √31 * i) / 2

This can be written as:

x = -3/2 ± (√31/2)i

Thus, the roots of the equation y = x² + 3x + 10 are:

  • x = -3/2 + (√31/2)i
  • x = -3/2 – (√31/2)i

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