What are the solutions to the equation -2x^2+8x-4=0 ?

The solutions to the equation -2x^2+8x-4=0 are x=2-sqrt(2),x=2+sqrt(2)

Find the zeros of -2x^2+8x-4=0

The zeros of -2x^2+8x-4=0 are x=2-sqrt(2),x=2+sqrt(2)

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- 2 x^{2} + 8 x - 4 = 0

x = 2 - 2, x = 2 + 2

Solution steps

- 2 x^{2} + 8 x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 2 ) ( - 4 )}{2 ( - 2 )}

8^{2} - 4 (- 2) (- 4) = 42

x_{1, 2} = \frac{- 8 \pm 4 2}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- 8 + 4 2}{2 ( - 2 )}, x_{2} = \frac{- 8 - 4 2}{2 ( - 2 )}

x = \frac{- 8 + 4 2}{2 ( - 2 )} : 2 - 2

x = \frac{- 8 - 4 2}{2 ( - 2 )} : 2 + 2

The solutions to the quadratic equation are:

x = 2 - 2, x = 2 + 2

4 x^{3} - 13 x^{2} + 4 x - 3

2 y^{2} + 11 y + 15

(\frac{27}{8})^{\frac{4}{3}}

x^{4} x^{10}

w^{2} + 12 w + 36

What are the solutions to the equation -2x^2+8x-4=0 ?
The solutions to the equation -2x^2+8x-4=0 are x=2-sqrt(2),x=2+sqrt(2)
Find the zeros of -2x^2+8x-4=0
The zeros of -2x^2+8x-4=0 are x=2-sqrt(2),x=2+sqrt(2)