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

The solutions to the equation 2x^2+4x-4=0 are x=-1+sqrt(3),x=-1-sqrt(3)

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

The zeros of 2x^2+4x-4=0 are x=-1+sqrt(3),x=-1-sqrt(3)

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

x = - 1 + 3, x = - 1 - 3

Decimal Notation

x = 0.73205 \dots, x = - 2.73205 \dots

Solution steps

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

Solve with the quadratic formula

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

4^{2} - 4 \cdot 2 (- 4) = 43

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

Separate the solutions

x_{1} = \frac{- 4 + 4 3}{2 \cdot 2}, x_{2} = \frac{- 4 - 4 3}{2 \cdot 2}

x = \frac{- 4 + 4 3}{2 \cdot 2} : - 1 + 3

x = \frac{- 4 - 4 3}{2 \cdot 2} : - 1 - 3

The solutions to the quadratic equation are:

x = - 1 + 3, x = - 1 - 3

y^{2} - 7 y - 18

2 l + 2

\frac{5.4 \cdot 1 0 ^{12}}{1.2 \cdot 1 0 ^{3}}

7 \cdot 16

x + 6 = 4

What are the solutions to the equation 2x^2+4x-4=0 ?
The solutions to the equation 2x^2+4x-4=0 are x=-1+sqrt(3),x=-1-sqrt(3)
Find the zeros of 2x^2+4x-4=0
The zeros of 2x^2+4x-4=0 are x=-1+sqrt(3),x=-1-sqrt(3)