What are the solutions to the equation -x(1-x)-6=0 ?

The solutions to the equation -x(1-x)-6=0 are x=3,x=-2

Find the zeros of -x(1-x)-6=0

The zeros of -x(1-x)-6=0 are x=3,x=-2

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- x (1 - x) - 6 = 0

x = 3, x = - 2

Solution steps

- x (1 - x) - 6 = 0

Expand

- x (1 - x) - 6 : - x + x^{2} - 6

- x + x^{2} - 6 = 0

Write in the standard form

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

x^{2} - x - 6 = 0

Solve with the quadratic formula

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

(- 1)^{2} - 4 \cdot 1 \cdot (- 6) = 5

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

Separate the solutions

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

x = \frac{- ( - 1 ) + 5}{2 \cdot 1} : 3

x = \frac{- ( - 1 ) - 5}{2 \cdot 1} : - 2

The solutions to the quadratic equation are:

x = 3, x = - 2

- 6 x^{2} + 12 = 0

60 x^{2} - 120 x = 0

34 x^{2} - 6 x = 0

x, (6 x - x^{2} y) = (x^{2} - 2 x)

1 2^{2} = x \times x + 7

What are the solutions to the equation -x(1-x)-6=0 ?
The solutions to the equation -x(1-x)-6=0 are x=3,x=-2
Find the zeros of -x(1-x)-6=0
The zeros of -x(1-x)-6=0 are x=3,x=-2