What are the solutions to the equation 2x^2-x+1=x+2 ?

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

Find the zeros of 2x^2-x+1=x+2

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

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

x = \frac{1 + 3}{2}, x = \frac{1 - 3}{2}

Solution steps

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

Move

2

to the left side

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

Move

x

to the left side

2 x^{2} - 2 x - 1 = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = \frac{1 + 3}{2}, x = \frac{1 - 3}{2}

c - 5 \leq 6

6 u - 19 = 3 u - 4

400 > (\frac{1}{20})^{7 x + 8}

3 x^{2} + 18 x + 33 = 6

\frac{- 4}{x ( x + 3 ) ( x - 1 ) ^{2}} \geq 0

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