What are the solutions to the equation n=x^2-2x ?

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

Find the zeros of n=x^2-2x

The zeros of n=x^2-2x are x=1+sqrt(n+1),x=1-sqrt(n+1)

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n = x^{2} - 2 x

x = 1 + n + 1, x = 1 - n + 1

Solution steps

n = x^{2} - 2 x

Switch sides

x^{2} - 2 x = n

Move

n

to the left side

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

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 1 + n + 1, x = 1 - n + 1

3 x^{2} - 32 x + 60 = 0

3 k^{2} - k = 0

3 k^{2} - k = 4

6 x^{2} + 48 x + 72 = 0

25 x^{2} + 60 x + 10 = 0

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