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

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

Find the zeros of roots x^2+4x-4

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

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roots

x^{2} + 4 x - 4

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

Solution steps

x^{2} + 4 x - 4

The roots are the intercepts with the x-axis

(y = 0)

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)

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

X^{2} = 81

x^{2} + 3 x - 6 = 0

s^{2} + 2 s - 3

s^{2} + 2 s + 6

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