What are the solutions to the equation 7x(1-x)=5(x-2) ?

The solutions to the equation 7x(1-x)=5(x-2) are x=-(-1+sqrt(71))/7 ,x=(1+sqrt(71))/7

Find the zeros of 7x(1-x)=5(x-2)

The zeros of 7x(1-x)=5(x-2) are x=-(-1+sqrt(71))/7 ,x=(1+sqrt(71))/7

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7 x (1 - x) = 5 (x - 2)

x = - \frac{- 1 + 71}{7}, x = \frac{1 + 71}{7}

Decimal Notation

x = - 1.06087 \dots, x = 1.34659 \dots

Solution steps

7 x (1 - x) = 5 (x - 2)

Expand

7 x (1 - x) : 7 x - 7 x^{2}

Expand

5 (x - 2) : 5 x - 10

7 x - 7 x^{2} = 5 x - 10

Move

10

to the left side

7 x - 7 x^{2} + 10 = 5 x

Move

5 x

to the left side

- 7 x^{2} + 2 x + 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 7 ) \cdot 10}{2 ( - 7 )}

2^{2} - 4 (- 7) \cdot 10 = 271

x_{1, 2} = \frac{- 2 \pm 2 71}{2 ( - 7 )}

Separate the solutions

x_{1} = \frac{- 2 + 2 71}{2 ( - 7 )}, x_{2} = \frac{- 2 - 2 71}{2 ( - 7 )}

x = \frac{- 2 + 2 71}{2 ( - 7 )} : - \frac{- 1 + 71}{7}

x = \frac{- 2 - 2 71}{2 ( - 7 )} : \frac{1 + 71}{7}

The solutions to the quadratic equation are:

x = - \frac{- 1 + 71}{7}, x = \frac{1 + 71}{7}

x^{2} - x - 2 = 0

x, x^{2} + 3 x - 2 = 0

(x + 4)^{2} + 50 = 0

k^{2} - 7 k + 6 = 0

(- 1 - x) (5 - x) + 9 = 0

What are the solutions to the equation 7x(1-x)=5(x-2) ?
The solutions to the equation 7x(1-x)=5(x-2) are x=-(-1+sqrt(71))/7 ,x=(1+sqrt(71))/7
Find the zeros of 7x(1-x)=5(x-2)
The zeros of 7x(1-x)=5(x-2) are x=-(-1+sqrt(71))/7 ,x=(1+sqrt(71))/7