What are the solutions to the equation -x^2+2x=3 ?

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

Find the zeros of -x^2+2x=3

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

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- x^{2} + 2 x = 3

x = 1 - 2 i, x = 1 + 2 i

Solution steps

- x^{2} + 2 x = 3

Move

3

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 1 ) ( - 3 )}{2 ( - 1 )}

Simplify

2^{2} - 4 (- 1) (- 3) : 22 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 1 - 2 i, x = 1 + 2 i

0.0057 v^{2} + 0.13 v - 100 = 0

v^{2} + 10 v + 24 = 0

(2.629 - x) (0.458 - x) - 0.80 1^{2} = 0

3 p^{2} = - 8 p - 3

8 - 6 m^{2} = - 88

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