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

The solutions to the equation x^2+12x+12=0 are x=2(sqrt(6)-3),x=-2(3+sqrt(6))

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

The zeros of x^2+12x+12=0 are x=2(sqrt(6)-3),x=-2(3+sqrt(6))

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

x = 2 (6 - 3), x = - 2 (3 + 6)

Solution steps

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

Solve with the quadratic formula

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

1 2^{2} - 4 \cdot 1 \cdot 12 = 46

x_{1, 2} = \frac{- 12 \pm 4 6}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 12 + 4 6}{2 \cdot 1}, x_{2} = \frac{- 12 - 4 6}{2 \cdot 1}

x = \frac{- 12 + 4 6}{2 \cdot 1} : 2 (6 - 3)

x = \frac{- 12 - 4 6}{2 \cdot 1} : - 2 (3 + 6)

The solutions to the quadratic equation are:

x = 2 (6 - 3), x = - 2 (3 + 6)

x^{2} - 5 x - 3 = 0

5 x - 3 (- x - 2) = 22

2 x^{2} + 7 x - 6 = 0

(\frac{1}{x})^{3}

7 n - 4 = 31

What are the solutions to the equation x^2+12x+12=0 ?
The solutions to the equation x^2+12x+12=0 are x=2(sqrt(6)-3),x=-2(3+sqrt(6))
Find the zeros of x^2+12x+12=0
The zeros of x^2+12x+12=0 are x=2(sqrt(6)-3),x=-2(3+sqrt(6))