Question

X^2+4x+5=0 ^2-квадрат

Answer 1

Решение:
${x}^{2} + 4x + 5 = 0 \\ d1 = {k}^{2} - ac = 4 - 5 < 0$
Ответ: Корней нет!

Answer 2

D=(4)^2-4*5*1=16-20=-4 , нет корней,вроде

Answer 3

The expression you provided, "X^2+4x+5=0," is a quadratic equation. To solve this equation, you can use the quadratic formula. The quadratic formula is given by:

x = (-b ± √(b^2 - 4ac)) / 2a

In your equation, a = 1, b = 4, and c = 5. Plug these values into the quadratic formula:

x = (-(4) ± √((4)^2 - 4 * 1 * 5)) / 2 * 1

Now, calculate the discriminant (the value inside the square root):

Discriminant (D) = (4)^2 - 4 * 1 * 5 = 16 - 20 = -4

Since the discriminant (D) is negative, the quadratic equation has no real solutions. Instead, it has two complex solutions. The solutions involve the imaginary unit 'i' (where i^2 = -1).

x = (-(4) ± √(-4)) / 2 * 1

x = (-4 ± 2i) / 2

Now, simplify each solution:

1. x = (-4 + 2i) / 2 x = -2 + i

2. x = (-4 - 2i) / 2 x = -2 - i

So, the solutions to the quadratic equation are x = -2 + i and x = -2 - i.