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

The solutions to the equation x^2+c=x are x=(1+sqrt(1-4c))/2 ,x=(1-sqrt(1-4c))/2

Find the zeros of x^2+c=x

The zeros of x^2+c=x are x=(1+sqrt(1-4c))/2 ,x=(1-sqrt(1-4c))/2

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x^{2} + c = x

x = \frac{1 + 1 - 4 c}{2}, x = \frac{1 - 1 - 4 c}{2}

Solution steps

x^{2} + c = x

Move

x

to the left side

x^{2} + c - x = 0

Write in the standard form

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

x^{2} - x + c = 0

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 1 - 4 c}{2 \cdot 1}, x_{2} = \frac{- ( - 1 ) - 1 - 4 c}{2 \cdot 1}

x = \frac{- ( - 1 ) + 1 - 4 c}{2 \cdot 1} : \frac{1 + 1 - 4 c}{2}

x = \frac{- ( - 1 ) - 1 - 4 c}{2 \cdot 1} : \frac{1 - 1 - 4 c}{2}

The solutions to the quadratic equation are:

x = \frac{1 + 1 - 4 c}{2}, x = \frac{1 - 1 - 4 c}{2}

2 x^{2} - 5 = 67

\frac{q ^{2}}{10000} - q + 3900 = 7000 - 1.1 q

x, x^{2} + 5 x - (p^{2} + p - 6) = 0

- 17 - 8 x^{2} = 4 x - 3 x^{2}

- x^{2} - 9 x + 22 = 0

What are the solutions to the equation x^2+c=x ?
The solutions to the equation x^2+c=x are x=(1+sqrt(1-4c))/2 ,x=(1-sqrt(1-4c))/2
Find the zeros of x^2+c=x
The zeros of x^2+c=x are x=(1+sqrt(1-4c))/2 ,x=(1-sqrt(1-4c))/2