What are the solutions to the equation n(n+1)=72 ?

The solutions to the equation n(n+1)=72 are n=8,n=-9

Find the zeros of n(n+1)=72

The zeros of n(n+1)=72 are n=8,n=-9

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n (n + 1) = 72

n = 8, n = - 9

Solution steps

n (n + 1) = 72

Expand

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

n^{2} + n = 72

Move

72

to the left side

n^{2} + n - 72 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 72 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 72) = 17

n_{1, 2} = \frac{- 1 \pm 17}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- 1 + 17}{2 \cdot 1}, n_{2} = \frac{- 1 - 17}{2 \cdot 1}

n = \frac{- 1 + 17}{2 \cdot 1} : 8

n = \frac{- 1 - 17}{2 \cdot 1} : - 9

The solutions to the quadratic equation are:

n = 8, n = - 9

z^{2} + z = 2

z^{2} + z = 1

4 (x + 12)^{2} = 28

x, 6 x^{2} + 2 y^{2} - 4 = 2 x^{2} y + 3 y

4 x^{2} + 7 x + 10 = 0

What are the solutions to the equation n(n+1)=72 ?
The solutions to the equation n(n+1)=72 are n=8,n=-9
Find the zeros of n(n+1)=72
The zeros of n(n+1)=72 are n=8,n=-9