What are the solutions to the equation x=x^2+2x-8 ?

The solutions to the equation x=x^2+2x-8 are x=(-1+sqrt(33))/2 ,x=(-1-sqrt(33))/2

Find the zeros of x=x^2+2x-8

The zeros of x=x^2+2x-8 are x=(-1+sqrt(33))/2 ,x=(-1-sqrt(33))/2

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x = x^{2} + 2 x - 8

x = \frac{- 1 + 33}{2}, x = \frac{- 1 - 33}{2}

Decimal Notation

x = 2.37228 \dots, x = - 3.37228 \dots

Solution steps

x = x^{2} + 2 x - 8

Switch sides

x^{2} + 2 x - 8 = x

Move

x

to the left side

x^{2} + x - 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 8 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 8) = 33

x_{1, 2} = \frac{- 1 \pm 33}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 1 + 33}{2 \cdot 1}, x_{2} = \frac{- 1 - 33}{2 \cdot 1}

x = \frac{- 1 + 33}{2 \cdot 1} : \frac{- 1 + 33}{2}

x = \frac{- 1 - 33}{2 \cdot 1} : \frac{- 1 - 33}{2}

The solutions to the quadratic equation are:

x = \frac{- 1 + 33}{2}, x = \frac{- 1 - 33}{2}

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

y^{2} - 10 y + 20 = 0

x, x^{2} + 2 x y + 3 y^{2} = 15

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

2 x^{2} + 16 x - 12 = 0

What are the solutions to the equation x=x^2+2x-8 ?
The solutions to the equation x=x^2+2x-8 are x=(-1+sqrt(33))/2 ,x=(-1-sqrt(33))/2
Find the zeros of x=x^2+2x-8
The zeros of x=x^2+2x-8 are x=(-1+sqrt(33))/2 ,x=(-1-sqrt(33))/2