6-x/x^2+2x+5>0

To solve the inequality \(\frac{6 - x}{x^2 + 2x + 5} > 0\), we need to analyze when the numerator and denominator are positive or negative.

Step 1: Analyze the denominator

First, let's examine the denominator \(x^2 + 2x + 5\):

• This is a quadratic expression. To check if it has any real roots, we can calculate the discriminant:

\[ \Delta = b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot 5 = 4 - 20 = -16 \]

Since the discriminant is negative, this quadratic has no real roots and opens upwards (as the coefficient of \(x^2\) is positive). Consequently, \(x^2 + 2x + 5 > 0\) for all real values of \(x\).

Step 2: Analyze the numerator

Next, let's analyze the numerator \(6 - x\):

• Set \(6 - x = 0\) to find where the numerator is zero:

\[ x = 6 \]

• The sign of \(6 - x\) changes at \(x = 6\):
  • For \(x < 6\), \(6 - x > 0\) (numerator is positive).
  • For \(x = 6\), \(6 - x = 0\) (numerator is zero).
  • For \(x > 6\), \(6 - x < 0\) (numerator is negative).

Step 3: Combine results

The fraction \(\frac{6 - x}{x^2 + 2x + 5}\) is positive when the numerator is positive and the denominator is positive. Because we found that the denominator is always positive, the sign of the overall expression is determined by the numerator.

• The inequality \(\frac{6 - x}{x^2 + 2x + 5} > 0\) holds true when:

\[ 6 - x > 0 \implies x < 6 \]

Conclusion

Thus, the solution to the inequality \(\frac{6 - x}{x^2 + 2x + 5} > 0\) is:

\[ \boxed{(-\infty, 6)} \]

This is the interval of all \(x\) values that satisfy the inequality.