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

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

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

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

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

x = 1 + 3, x = 1 - 3

Solution steps

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

Switch sides

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

(1 + \frac{0.08}{4})^{4} - 1

(x + 3) (3 x - 2) = (4 x + 5) (2 x - 3)

(x - 3)^{2} = - 36

2 x (4 x^{2})

t^{4} - 16

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