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 \cdot 1 \cdot 2}{2 \cdot 1}

Simplify

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 1 + i, x = 1 - i

2 ∣ 3 x - 2 ∣ \geq 10

4 x^{2} + \frac{5}{3} = 0

4 z^{2} + 3 z - 1 = 0

2 < x - 2 < 7

y^{\frac{3}{4}} \cdot y^{\frac{1}{3}}

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