What are the solutions to the equation n^2-2n=0 ?

The solutions to the equation n^2-2n=0 are n=2,n=0

Find the zeros of n^2-2n=0

The zeros of n^2-2n=0 are n=2,n=0

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n^{2} - 2 n = 0

n = 2, n = 0

Solution steps

n^{2} - 2 n = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot 0 = 2

n_{1, 2} = \frac{- ( - 2 ) \pm 2}{2 \cdot 1}

Separate the solutions

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

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

n = \frac{- ( - 2 ) - 2}{2 \cdot 1} : 0

The solutions to the quadratic equation are:

n = 2, n = 0

2 x^{2} + 4 x - 2

\frac{a + 2}{6 b} + \frac{7 b}{3 a + 6}

\frac{5}{7} (x - 3) \geq 4 x + \frac{22}{7}

6 x^{3} + 3 x \geq 4 x^{2} + 2

(- 2 r^{3} p^{2} m)^{3}

What are the solutions to the equation n^2-2n=0 ?
The solutions to the equation n^2-2n=0 are n=2,n=0
Find the zeros of n^2-2n=0
The zeros of n^2-2n=0 are n=2,n=0