What are the solutions to the equation 3x^2+12x+28=0 ?

The solutions to the equation 3x^2+12x+28=0 are x=-2+i(4sqrt(3))/3 ,x=-2-i(4sqrt(3))/3

Find the zeros of 3x^2+12x+28=0

The zeros of 3x^2+12x+28=0 are x=-2+i(4sqrt(3))/3 ,x=-2-i(4sqrt(3))/3

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3 x^{2} + 12 x + 28 = 0

x = - 2 + i \frac{4 3}{3}, x = - 2 - i \frac{4 3}{3}

Solution steps

3 x^{2} + 12 x + 28 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 3 \cdot 28}{2 \cdot 3}

Simplify

1 2^{2} - 4 \cdot 3 \cdot 28 : 83 i

x_{1, 2} = \frac{- 12 \pm 8 3 i}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 12 + 8 3 i}{2 \cdot 3}, x_{2} = \frac{- 12 - 8 3 i}{2 \cdot 3}

x = \frac{- 12 + 8 3 i}{2 \cdot 3} : - 2 + i \frac{4 3}{3}

x = \frac{- 12 - 8 3 i}{2 \cdot 3} : - 2 - i \frac{4 3}{3}

The solutions to the quadratic equation are:

x = - 2 + i \frac{4 3}{3}, x = - 2 - i \frac{4 3}{3}

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

2 (x - 6)^{2} + 5 = 23

(x + 3)^{2} - 3 = 9

5 (2 n - 1)^{2} = 45

2 x^{2} + 4 x = 8

What are the solutions to the equation 3x^2+12x+28=0 ?
The solutions to the equation 3x^2+12x+28=0 are x=-2+i(4sqrt(3))/3 ,x=-2-i(4sqrt(3))/3
Find the zeros of 3x^2+12x+28=0
The zeros of 3x^2+12x+28=0 are x=-2+i(4sqrt(3))/3 ,x=-2-i(4sqrt(3))/3