What are the solutions to the equation n(n+1)=2 ?

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

Find the zeros of n(n+1)=2

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

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n (n + 1) = 2

n = 1, n = - 2

Solution steps

n (n + 1) = 2

Expand

n (n + 1) : n^{2} + n

n^{2} + n = 2

Move

2

to the left side

n^{2} + n - 2 = 0

Solve with the quadratic formula

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

1^{2} - 4 \cdot 1 \cdot (- 2) = 3

n_{1, 2} = \frac{- 1 \pm 3}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- 1 + 3}{2 \cdot 1}, n_{2} = \frac{- 1 - 3}{2 \cdot 1}

n = \frac{- 1 + 3}{2 \cdot 1} : 1

n = \frac{- 1 - 3}{2 \cdot 1} : - 2

The solutions to the quadratic equation are:

n = 1, n = - 2

- 8 z = 12 + z^{2}

(x - 1)^{2} - (x - 1) = 42

x^{2} + 11 x - 17 = 0

5 x^{2} - 2 = - 9 x

y^{2} + 3 y = 2

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