What are the solutions to the equation x=2x^2+3x-1 ?

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

Find the zeros of x=2x^2+3x-1

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

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

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

Decimal Notation

x = 0.36602 \dots, x = - 1.36602 \dots

Solution steps

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

Switch sides

2 x^{2} + 3 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}

3 x^{2} + 15 = 18 x

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

2 x^{2} - 3 x - 15 = 0

12 x^{2} + x = 63

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

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