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

The solutions to the equation x^2-2(2-x)=2 are x=-1+sqrt(7),x=-1-sqrt(7)

Find the zeros of x^2-2(2-x)=2

The zeros of x^2-2(2-x)=2 are x=-1+sqrt(7),x=-1-sqrt(7)

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x^{2} - 2 (2 - x) = 2

x = - 1 + 7, x = - 1 - 7

Decimal Notation

x = 1.64575 \dots, x = - 3.64575 \dots

Solution steps

x^{2} - 2 (2 - x) = 2

Expand

x^{2} - 2 (2 - x) : x^{2} - 4 + 2 x

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

Move

2

to the left side

x^{2} + 2 x - 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 6 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 6) = 27

x_{1, 2} = \frac{- 2 \pm 2 7}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 2 + 2 7}{2 \cdot 1}, x_{2} = \frac{- 2 - 2 7}{2 \cdot 1}

x = \frac{- 2 + 2 7}{2 \cdot 1} : - 1 + 7

x = \frac{- 2 - 2 7}{2 \cdot 1} : - 1 - 7

The solutions to the quadratic equation are:

x = - 1 + 7, x = - 1 - 7

12 x^{2} - 9 x = 0

5 x + \frac{4 + 4 x}{2} = 23

2 n - 1 - 3 n = 5 n - 7 - 3 n

\frac{4 ( 3 x - 2 )}{3} - \frac{5 ( 8 x - 5 )}{4} = \frac{5}{3}

2 x + 3 (x - 4) = 2 (x - 3)

What are the solutions to the equation x^2-2(2-x)=2 ?
The solutions to the equation x^2-2(2-x)=2 are x=-1+sqrt(7),x=-1-sqrt(7)
Find the zeros of x^2-2(2-x)=2
The zeros of x^2-2(2-x)=2 are x=-1+sqrt(7),x=-1-sqrt(7)