What are the solutions to the equation x^2+5=x ?

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

Find the zeros of x^2+5=x

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

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x^{2} + 5 = x

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

Solution steps

x^{2} + 5 = x

Move

x

to the left side

x^{2} + 5 - x = 0

Write in the standard form

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

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}

2 x^{2} = 14 x

2 x^{2} + 4 x + 1 = 0

t, f = 29 - (x^{2} + z^{2} + t^{2})

3 x^{2} - 6 x + 10

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

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