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

The solutions to the equation -2x^2+4x+8=0 are x=1-sqrt(5),x=1+sqrt(5)

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

The zeros of -2x^2+4x+8=0 are x=1-sqrt(5),x=1+sqrt(5)

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

x = 1 - 5, x = 1 + 5

Decimal Notation

x = - 1.23606 \dots, x = 3.23606 \dots

Solution steps

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

Solve with the quadratic formula

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

4^{2} - 4 (- 2) \cdot 8 = 45

x_{1, 2} = \frac{- 4 \pm 4 5}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- 4 + 4 5}{2 ( - 2 )}, x_{2} = \frac{- 4 - 4 5}{2 ( - 2 )}

x = \frac{- 4 + 4 5}{2 ( - 2 )} : 1 - 5

x = \frac{- 4 - 4 5}{2 ( - 2 )} : 1 + 5

The solutions to the quadratic equation are:

x = 1 - 5, x = 1 + 5

3 x^{2} - (- 4 x + 7) = 2 x^{2} + 5 x - 9

x, - y (x^{2} - y) - (x + 1) (x - 1) = 0

3 w^{2} - w + 5 = 6

x^{2} - 3 x - 1 = 0

x^{2} = 3

What are the solutions to the equation -2x^2+4x+8=0 ?
The solutions to the equation -2x^2+4x+8=0 are x=1-sqrt(5),x=1+sqrt(5)
Find the zeros of -2x^2+4x+8=0
The zeros of -2x^2+4x+8=0 are x=1-sqrt(5),x=1+sqrt(5)