What are the solutions to the equation x^2-x-2=(1-x)/2 ?

The solutions to the equation x^2-x-2=(1-x)/2 are x=(1+sqrt(41))/4 ,x=(1-sqrt(41))/4

Find the zeros of x^2-x-2=(1-x)/2

The zeros of x^2-x-2=(1-x)/2 are x=(1+sqrt(41))/4 ,x=(1-sqrt(41))/4

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

x = \frac{1 + 41}{4}, x = \frac{1 - 41}{4}

Decimal Notation

x = 1.85078 \dots, x = - 1.35078 \dots

Solution steps

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

Multiply both sides by

2

2 x^{2} - 2 x - 4 = 1 - x

Move

x

to the left side

2 x^{2} - x - 4 = 1

Move

1

to the left side

2 x^{2} - x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 2 ( - 5 )}{2 \cdot 2}

(- 1)^{2} - 4 \cdot 2 (- 5) = 41

x_{1, 2} = \frac{- ( - 1 ) \pm 41}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 41}{2 \cdot 2}, x_{2} = \frac{- ( - 1 ) - 41}{2 \cdot 2}

x = \frac{- ( - 1 ) + 41}{2 \cdot 2} : \frac{1 + 41}{4}

x = \frac{- ( - 1 ) - 41}{2 \cdot 2} : \frac{1 - 41}{4}

The solutions to the quadratic equation are:

x = \frac{1 + 41}{4}, x = \frac{1 - 41}{4}

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(4 x^{4} y^{- 4})^{3}

(2 + 8 i) (2 - 8 i)

What are the solutions to the equation x^2-x-2=(1-x)/2 ?
The solutions to the equation x^2-x-2=(1-x)/2 are x=(1+sqrt(41))/4 ,x=(1-sqrt(41))/4
Find the zeros of x^2-x-2=(1-x)/2
The zeros of x^2-x-2=(1-x)/2 are x=(1+sqrt(41))/4 ,x=(1-sqrt(41))/4