What are the solutions to the equation 2x^2+4x+12=0 ?

The solutions to the equation 2x^2+4x+12=0 are x=-1+sqrt(5)i,x=-1-sqrt(5)i

Find the zeros of 2x^2+4x+12=0

The zeros of 2x^2+4x+12=0 are x=-1+sqrt(5)i,x=-1-sqrt(5)i

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2 x^{2} + 4 x + 12 = 0

x = - 1 + 5 i, x = - 1 - 5 i

Solution steps

2 x^{2} + 4 x + 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 2 \cdot 12}{2 \cdot 2}

Simplify

4^{2} - 4 \cdot 2 \cdot 12 : 45 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = - 1 + 5 i, x = - 1 - 5 i

3 p^{2} - 75

x^{2} - 2 x - 16 = 8

5 (x + 1) = 3 x + 2 (x + 1) + 4

- 1 \leq 4 x + 3 < 11

x^{3} - 7 x - 6 > 0

What are the solutions to the equation 2x^2+4x+12=0 ?
The solutions to the equation 2x^2+4x+12=0 are x=-1+sqrt(5)i,x=-1-sqrt(5)i
Find the zeros of 2x^2+4x+12=0
The zeros of 2x^2+4x+12=0 are x=-1+sqrt(5)i,x=-1-sqrt(5)i