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

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

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

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

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

x = 1 - 3 i, x = 1 + 3 i

Solution steps

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

Solve with the quadratic formula

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

Simplify

2^{2} - 4 (- 1) (- 4) : 23 i

x_{1, 2} = \frac{- 2 \pm 2 3 i}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 2 + 2 3 i}{2 ( - 1 )}, x_{2} = \frac{- 2 - 2 3 i}{2 ( - 1 )}

x = \frac{- 2 + 2 3 i}{2 ( - 1 )} : 1 - 3 i

x = \frac{- 2 - 2 3 i}{2 ( - 1 )} : 1 + 3 i

The solutions to the quadratic equation are:

x = 1 - 3 i, x = 1 + 3 i

2 f + 3 = 11 f - 24

x^{2} - 121 \geq 0

- 3 + 2 b < 13

1331

- 20.32 = 6.85 - 9.81 (t)

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