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

The solutions to the equation solvefor x,2x^2+12x+4=0 are x=-3+sqrt(7),x=-3-sqrt(7)

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

The zeros of solvefor x,2x^2+12x+4=0 are x=-3+sqrt(7),x=-3-sqrt(7)

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

x = - 3 + 7, x = - 3 - 7

Decimal Notation

x = - 0.35424 \dots, x = - 5.64575 \dots

Solution steps

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

Solve with the quadratic formula

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

1 2^{2} - 4 \cdot 2 \cdot 4 = 47

x_{1, 2} = \frac{- 12 \pm 4 7}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 12 + 4 7}{2 \cdot 2}, x_{2} = \frac{- 12 - 4 7}{2 \cdot 2}

x = \frac{- 12 + 4 7}{2 \cdot 2} : - 3 + 7

x = \frac{- 12 - 4 7}{2 \cdot 2} : - 3 - 7

The solutions to the quadratic equation are:

x = - 3 + 7, x = - 3 - 7

m^{2} + 40 m + 4000 = 0

c^{2} + 10 c + 8 = 0

4 z^{2} - 5 z + 3 = 0

x, x^{2} + 2 x y - y^{2} = 0

x + 1 = (x + 1)^{2}

What are the solutions to the equation solvefor x,2x^2+12x+4=0 ?
The solutions to the equation solvefor x,2x^2+12x+4=0 are x=-3+sqrt(7),x=-3-sqrt(7)
Find the zeros of solvefor x,2x^2+12x+4=0
The zeros of solvefor x,2x^2+12x+4=0 are x=-3+sqrt(7),x=-3-sqrt(7)