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

The solutions to the equation -x+2=x^2-2 are x=(-1+sqrt(17))/2 ,x=(-1-sqrt(17))/2

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

The zeros of -x+2=x^2-2 are x=(-1+sqrt(17))/2 ,x=(-1-sqrt(17))/2

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

x = \frac{- 1 + 17}{2}, x = \frac{- 1 - 17}{2}

Decimal Notation

x = 1.56155 \dots, x = - 2.56155 \dots

Solution steps

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

Switch sides

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

Move

2

to the left side

x^{2} - 4 = - x

Move

x

to the left side

x^{2} - 4 + x = 0

Write in the standard form

a x^{2} + b x + c = 0

x^{2} + x - 4 = 0

Solve with the quadratic formula

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

1^{2} - 4 \cdot 1 \cdot (- 4) = 17

x_{1, 2} = \frac{- 1 \pm 17}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 1 + 17}{2 \cdot 1}, x_{2} = \frac{- 1 - 17}{2 \cdot 1}

x = \frac{- 1 + 17}{2 \cdot 1} : \frac{- 1 + 17}{2}

x = \frac{- 1 - 17}{2 \cdot 1} : \frac{- 1 - 17}{2}

The solutions to the quadratic equation are:

x = \frac{- 1 + 17}{2}, x = \frac{- 1 - 17}{2}

5 v^{2} + 10 v = 0

v, e = \frac{m v ^{2}}{2} + m g h

7 x^{2} + 9 x - 148 = 0

x, (5 x + y)^{2} + 9 x - 2 y = 4 x (- 8 y + 3)^{3}

(x - 3)^{2} - 9 = - x

What are the solutions to the equation -x+2=x^2-2 ?
The solutions to the equation -x+2=x^2-2 are x=(-1+sqrt(17))/2 ,x=(-1-sqrt(17))/2
Find the zeros of -x+2=x^2-2
The zeros of -x+2=x^2-2 are x=(-1+sqrt(17))/2 ,x=(-1-sqrt(17))/2