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

The solutions to the equation -x+4x(2-x)=-1 are x=-(-7+sqrt(65))/8 ,x=(7+sqrt(65))/8

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

The zeros of -x+4x(2-x)=-1 are x=-(-7+sqrt(65))/8 ,x=(7+sqrt(65))/8

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- x + 4 x (2 - x) = - 1

x = - \frac{- 7 + 65}{8}, x = \frac{7 + 65}{8}

Decimal Notation

x = - 0.13278 \dots, x = 1.88278 \dots

Solution steps

- x + 4 x (2 - x) = - 1

Expand

- x + 4 x (2 - x) : 7 x - 4 x^{2}

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

Move

1

to the left side

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

Write in the standard form

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

- 4 x^{2} + 7 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 ( - 4 ) \cdot 1}{2 ( - 4 )}

7^{2} - 4 (- 4) \cdot 1 = 65

x_{1, 2} = \frac{- 7 \pm 65}{2 ( - 4 )}

Separate the solutions

x_{1} = \frac{- 7 + 65}{2 ( - 4 )}, x_{2} = \frac{- 7 - 65}{2 ( - 4 )}

x = \frac{- 7 + 65}{2 ( - 4 )} : - \frac{- 7 + 65}{8}

x = \frac{- 7 - 65}{2 ( - 4 )} : \frac{7 + 65}{8}

The solutions to the quadratic equation are:

x = - \frac{- 7 + 65}{8}, x = \frac{7 + 65}{8}

l (x) = - x^{2} + 12 x - 20

x^{2} + 1.4 = 3.6

12 h^{2} - h - 6 = 0

3 x^{2} - 5 = 34

6 x^{2} + x = 0

What are the solutions to the equation -x+4x(2-x)=-1 ?
The solutions to the equation -x+4x(2-x)=-1 are x=-(-7+sqrt(65))/8 ,x=(7+sqrt(65))/8
Find the zeros of -x+4x(2-x)=-1
The zeros of -x+4x(2-x)=-1 are x=-(-7+sqrt(65))/8 ,x=(7+sqrt(65))/8