What are the solutions to the equation x^2+4x+24=0 ?

The solutions to the equation x^2+4x+24=0 are x=-2+2sqrt(5)i,x=-2-2sqrt(5)i

Find the zeros of x^2+4x+24=0

The zeros of x^2+4x+24=0 are x=-2+2sqrt(5)i,x=-2-2sqrt(5)i

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x^{2} + 4 x + 24 = 0

x = - 2 + 25 i, x = - 2 - 25 i

Solution steps

x^{2} + 4 x + 24 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot 24}{2 \cdot 1}

Simplify

4^{2} - 4 \cdot 1 \cdot 24 : 45 i

x_{1, 2} = \frac{- 4 \pm 4 5 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 4 + 4 5 i}{2 \cdot 1}, x_{2} = \frac{- 4 - 4 5 i}{2 \cdot 1}

x = \frac{- 4 + 4 5 i}{2 \cdot 1} : - 2 + 25 i

x = \frac{- 4 - 4 5 i}{2 \cdot 1} : - 2 - 25 i

The solutions to the quadratic equation are:

x = - 2 + 25 i, x = - 2 - 25 i

4 x^{2} - 6 = 30

x, y = 0.5 (x - 3) (x + k)

x^{2} + 14 x + 49 = 25

(x - 8) (2 x + 5) = 0

3 x^{2} - 7 x - 10 = 0

What are the solutions to the equation x^2+4x+24=0 ?
The solutions to the equation x^2+4x+24=0 are x=-2+2sqrt(5)i,x=-2-2sqrt(5)i
Find the zeros of x^2+4x+24=0
The zeros of x^2+4x+24=0 are x=-2+2sqrt(5)i,x=-2-2sqrt(5)i