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

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

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

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

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x (x - 1) = - 1

x = \frac{1}{2} + i \frac{3}{2}, x = \frac{1}{2} - i \frac{3}{2}

Solution steps

x (x - 1) = - 1

Expand

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

x^{2} - x = - 1

Move

1

to the left side

x^{2} - x + 1 = 0

Solve with the quadratic formula

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

Simplify

(- 1)^{2} - 4 \cdot 1 \cdot 1 : 3 i

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

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 3 i}{2 \cdot 1}, x_{2} = \frac{- ( - 1 ) - 3 i}{2 \cdot 1}

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

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

The solutions to the quadratic equation are:

x = \frac{1}{2} + i \frac{3}{2}, x = \frac{1}{2} - i \frac{3}{2}

4 (t - 1)^{2} - 9 (t - 1) = - 2

x^{2} = - 6 x - 5

a x^{2} - 3 x - 5 = 0

25 v^{2} - 36 = 0

2 x^{2} - 24 x + a^{3} = 0

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