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

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

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

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

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

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

Decimal Notation

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

Solution steps

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

Switch sides

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 (1 + 2)

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

The solutions to the quadratic equation are:

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

(\frac{5}{2})^{- 2} \cdot \frac{25}{8}

1 - \frac{0.05}{2}

x^{2} + 12 x - 28 = 0

\frac{x ^{- 8}}{x ^{- 7}}

(x - 2)^{2} - 4 (x - 2) - 21 = 0

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