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

The solutions to the equation -x^2+2x=2 are x=1-i,x=1+i

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

The zeros of -x^2+2x=2 are x=1-i,x=1+i

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- x^{2} + 2 x = 2

x = 1 - i, x = 1 + i

Solution steps

- x^{2} + 2 x = 2

Move

2

to the left side

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

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 1 - i, x = 1 + i

8 x^{2} + 7 = 5 x^{2} + 4

7 t (t + (17)) = - 420

6 x^{2} + 45 x + 21 = 0

x^{2} + 12 x + c

- x (- 6 (x) + 1) = 1

What are the solutions to the equation -x^2+2x=2 ?
The solutions to the equation -x^2+2x=2 are x=1-i,x=1+i
Find the zeros of -x^2+2x=2
The zeros of -x^2+2x=2 are x=1-i,x=1+i