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

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

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

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

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

x = 1 + 3, x = 1 - 3

Decimal Notation

x = 2.73205 \dots, x = - 0.73205 \dots

Solution steps

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

Move

4

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}

(- 2)^{2} - 4 \cdot 1 \cdot (- 2) = 23

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 1 + 3, x = 1 - 3

4 x^{2} + 12 x = 3

20 w^{2} + 13 w + 2 = 0

\frac{x ^{2} - 1}{2} - \frac{3 x - 1}{4} = 2

81 x^{2} + 36 x = - 4

x, m x^{2} + 6 x - 3 = 0

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