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

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

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

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

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

x = 2 (2 - 1), x = - 2 (1 + 2)

Decimal Notation

x = 0.82842 \dots, x = - 4.82842 \dots

Solution steps

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

Solve with the quadratic formula

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

4^{2} - 4 \cdot 1 \cdot (- 4) = 42

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 2 (2 - 1), x = - 2 (1 + 2)

\frac{x}{- 7} = - 2 - (- 8)

27 x^{3} + y^{6}

\frac{11}{20} - \frac{1}{28}

(32 x^{25} y^{- 10})^{\frac{2}{5}}

0.06 x + 0.07 (38000 - x) = 2440

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