What are the solutions to the equation x-x^2=1-6x^2 ?

The solutions to the equation x-x^2=1-6x^2 are x=(-1+sqrt(21))/(10),x=(-1-sqrt(21))/(10)

Find the zeros of x-x^2=1-6x^2

The zeros of x-x^2=1-6x^2 are x=(-1+sqrt(21))/(10),x=(-1-sqrt(21))/(10)

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

x = \frac{- 1 + 21}{10}, x = \frac{- 1 - 21}{10}

Decimal Notation

x = 0.35825 \dots, x = - 0.55825 \dots

Solution steps

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

Move

6 x^{2}

to the left side

x + 5 x^{2} = 1

Move

1

to the left side

x + 5 x^{2} - 1 = 0

Write in the standard form

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

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 5 ( - 1 )}{2 \cdot 5}

1^{2} - 4 \cdot 5 (- 1) = 21

x_{1, 2} = \frac{- 1 \pm 21}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- 1 + 21}{2 \cdot 5}, x_{2} = \frac{- 1 - 21}{2 \cdot 5}

x = \frac{- 1 + 21}{2 \cdot 5} : \frac{- 1 + 21}{10}

x = \frac{- 1 - 21}{2 \cdot 5} : \frac{- 1 - 21}{10}

The solutions to the quadratic equation are:

x = \frac{- 1 + 21}{10}, x = \frac{- 1 - 21}{10}

(8 \times 1 0^{- 3}) (2.71775 - 1950 x)^{2} - x = 0

16 p^{2} - 49

\frac{- 4}{4}

\frac{y ^{- 5}}{y ^{- 2}}

\frac{- 3}{\frac{5}{x} + y}

What are the solutions to the equation x-x^2=1-6x^2 ?
The solutions to the equation x-x^2=1-6x^2 are x=(-1+sqrt(21))/(10),x=(-1-sqrt(21))/(10)
Find the zeros of x-x^2=1-6x^2
The zeros of x-x^2=1-6x^2 are x=(-1+sqrt(21))/(10),x=(-1-sqrt(21))/(10)