What are the solutions to the equation (x-1)^2+(x-1)+5=0 ?

The solutions to the equation (x-1)^2+(x-1)+5=0 are x= 1/2+i(sqrt(19))/2 ,x= 1/2-i(sqrt(19))/2

Find the zeros of (x-1)^2+(x-1)+5=0

The zeros of (x-1)^2+(x-1)+5=0 are x= 1/2+i(sqrt(19))/2 ,x= 1/2-i(sqrt(19))/2

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(x - 1)^{2} + (x - 1) + 5 = 0

x = \frac{1}{2} + i \frac{19}{2}, x = \frac{1}{2} - i \frac{19}{2}

Solution steps

(x - 1)^{2} + (x - 1) + 5 = 0

Expand

(x - 1)^{2} + (x - 1) + 5 : x^{2} - x + 5

x^{2} - x + 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot 5}{2 \cdot 1}

Simplify

(- 1)^{2} - 4 \cdot 1 \cdot 5 : 19 i

x_{1, 2} = \frac{- ( - 1 ) \pm 19 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 19 i}{2 \cdot 1}, x_{2} = \frac{- ( - 1 ) - 19 i}{2 \cdot 1}

x = \frac{- ( - 1 ) + 19 i}{2 \cdot 1} : \frac{1}{2} + i \frac{19}{2}

x = \frac{- ( - 1 ) - 19 i}{2 \cdot 1} : \frac{1}{2} - i \frac{19}{2}

The solutions to the quadratic equation are:

x = \frac{1}{2} + i \frac{19}{2}, x = \frac{1}{2} - i \frac{19}{2}

(x + a)^{2} = b

h^{2} + 5 h - 204 = 0

- 4.9 t^{2} + 19.6 t + 58.8 = 0

x^{2} - 2 x - 36 = 0

4 x^{2} + 36 x + 85 = 0

What are the solutions to the equation (x-1)^2+(x-1)+5=0 ?
The solutions to the equation (x-1)^2+(x-1)+5=0 are x= 1/2+i(sqrt(19))/2 ,x= 1/2-i(sqrt(19))/2
Find the zeros of (x-1)^2+(x-1)+5=0
The zeros of (x-1)^2+(x-1)+5=0 are x= 1/2+i(sqrt(19))/2 ,x= 1/2-i(sqrt(19))/2