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

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

Find the zeros of 2x(x+1)=1

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

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2 x (x + 1) = 1

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

Decimal Notation

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

Solution steps

2 x (x + 1) = 1

Expand

2 x (x + 1) : 2 x^{2} + 2 x

2 x^{2} + 2 x = 1

Move

1

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 n^{2} - 8 n - 3 = 0

9 m^{3} + 6 m

\frac{x ^{2} y}{x - 9} \cdot \frac{x ^{2} - 81}{x y ^{2}}

3 < - 5 x + 8 < 10

y^{\frac{3}{4}} \cdot y^{\frac{1}{3}}

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