how many solutions are there for f(x) = x^2 + 4x + 5?

Answer 1

The function f(x) = x^2 + 4x + 5 is a quadratic function, and the solutions can be found by setting the function equal to zero and solving for x.

To find the solutions for f(x) = x^2 + 4x + 5, we set the function equal to zero:

x^2 + 4x + 5 = 0

Next, we can use the quadratic formula to find the solutions:

x = [-b ± sqrt(b^2 - 4ac)] / 2a

In this case, a = 1, b = 4, and c = 5. Substituting these values into the formula, we get:

x = [-4 ± sqrt(4^2 - 4(1)(5))] / (2*1)
x = [-4 ± sqrt(16 - 20)] / 2
x = [-4 ± sqrt(-4)] / 2
x = [-4 ± 2i] / 2
x = -2 ± i

Therefore, the function f(x) = x^2 + 4x + 5 has two complex solutions: -2 + i and -2 - i.