What are the solutions to the equation 2x(1-x)=-3 ?

The solutions to the equation 2x(1-x)=-3 are x=-(-1+sqrt(7))/2 ,x=(1+sqrt(7))/2

Find the zeros of 2x(1-x)=-3

The zeros of 2x(1-x)=-3 are x=-(-1+sqrt(7))/2 ,x=(1+sqrt(7))/2

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2 x (1 - x) = - 3

x = - \frac{- 1 + 7}{2}, x = \frac{1 + 7}{2}

Decimal Notation

x = - 0.82287 \dots, x = 1.82287 \dots

Solution steps

2 x (1 - x) = - 3

Expand

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

2 x - 2 x^{2} = - 3

Move

3

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 2 ) \cdot 3}{2 ( - 2 )}

2^{2} - 4 (- 2) \cdot 3 = 27

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = - \frac{- 1 + 7}{2}, x = \frac{1 + 7}{2}

- 3 x^{2} + 12 x - 15 = 0

- 2 x^{2} - 5 = 0

9 x^{2} + 2 x = 0

- 20 v - 25 = 4 v^{2}

t^{2} - 14 t

What are the solutions to the equation 2x(1-x)=-3 ?
The solutions to the equation 2x(1-x)=-3 are x=-(-1+sqrt(7))/2 ,x=(1+sqrt(7))/2
Find the zeros of 2x(1-x)=-3
The zeros of 2x(1-x)=-3 are x=-(-1+sqrt(7))/2 ,x=(1+sqrt(7))/2