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

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

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

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

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(x + 2) \cdot 2 x + (x - 1)^{2} = 2

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

Solution steps

(x + 2) \cdot 2 x + (x - 1)^{2} = 2

Expand

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

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

Move

2

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 3 ( - 1 )}{2 \cdot 3}

2^{2} - 4 \cdot 3 (- 1) = 4

x_{1, 2} = \frac{- 2 \pm 4}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 2 + 4}{2 \cdot 3}, x_{2} = \frac{- 2 - 4}{2 \cdot 3}

x = \frac{- 2 + 4}{2 \cdot 3} : \frac{1}{3}

x = \frac{- 2 - 4}{2 \cdot 3} : - 1

The solutions to the quadratic equation are:

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

m x + m

2 y + 6 = 1 - 3 y

x^{2} - 8 x > - 12

2 (x + 3) = 10

3 x + x + \frac{2}{3} x = 28

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