What are the solutions to the equation -2x+4(2x+x^2)=-1 ?

The solutions to the equation -2x+4(2x+x^2)=-1 are x=(-3+sqrt(5))/4 ,x=-(3+sqrt(5))/4

Find the zeros of -2x+4(2x+x^2)=-1

The zeros of -2x+4(2x+x^2)=-1 are x=(-3+sqrt(5))/4 ,x=-(3+sqrt(5))/4

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

x = \frac{- 3 + 5}{4}, x = - \frac{3 + 5}{4}

Decimal Notation

x = - 0.19098 \dots, x = - 1.30901 \dots

Solution steps

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

Expand

- 2 x + 4 (2 x + x^{2}) : 6 x + 4 x^{2}

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

Move

1

to the left side

6 x + 4 x^{2} + 1 = 0

Write in the standard form

a x^{2} + b x + c = 0

4 x^{2} + 6 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 4 \cdot 1}{2 \cdot 4}

6^{2} - 4 \cdot 4 \cdot 1 = 25

x_{1, 2} = \frac{- 6 \pm 2 5}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 6 + 2 5}{2 \cdot 4}, x_{2} = \frac{- 6 - 2 5}{2 \cdot 4}

x = \frac{- 6 + 2 5}{2 \cdot 4} : \frac{- 3 + 5}{4}

x = \frac{- 6 - 2 5}{2 \cdot 4} : - \frac{3 + 5}{4}

The solutions to the quadratic equation are:

x = \frac{- 3 + 5}{4}, x = - \frac{3 + 5}{4}

x^{2} - 12 x + 17 = 0

d^{2} - 10 d + 25 = 0

0 = 2 x^{2} - 4 x - 1

9 x^{2} + 36 x + 72 = 0

x, y^{2} + x^{2} = 0

What are the solutions to the equation -2x+4(2x+x^2)=-1 ?
The solutions to the equation -2x+4(2x+x^2)=-1 are x=(-3+sqrt(5))/4 ,x=-(3+sqrt(5))/4
Find the zeros of -2x+4(2x+x^2)=-1
The zeros of -2x+4(2x+x^2)=-1 are x=(-3+sqrt(5))/4 ,x=-(3+sqrt(5))/4