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 ( - 1 ) \cdot 4}{2 ( - 1 )}

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

45 + 20 - 5

x^{- 2}

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

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

\frac{3 x + 6}{x - 5} > 0

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))