What are the solutions to the equation n(n+1)=240 ?

The solutions to the equation n(n+1)=240 are n=15,n=-16

Find the zeros of n(n+1)=240

The zeros of n(n+1)=240 are n=15,n=-16

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n (n + 1) = 240

n = 15, n = - 16

Solution steps

n (n + 1) = 240

Expand

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

n^{2} + n = 240

Move

240

to the left side

n^{2} + n - 240 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 240 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 240) = 31

n_{1, 2} = \frac{- 1 \pm 31}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- 1 + 31}{2 \cdot 1}, n_{2} = \frac{- 1 - 31}{2 \cdot 1}

n = \frac{- 1 + 31}{2 \cdot 1} : 15

n = \frac{- 1 - 31}{2 \cdot 1} : - 16

The solutions to the quadratic equation are:

n = 15, n = - 16

- 0.06 x^{2} + 4 x - 35 = 0

16 9^{2} = 12 5^{2} + x^{2}

(x - 2)^{2} = 5 - (x - 1)^{2}

4.7 x^{2} + 9 x = 0

x^{2} + 4 x = 100

What are the solutions to the equation n(n+1)=240 ?
The solutions to the equation n(n+1)=240 are n=15,n=-16
Find the zeros of n(n+1)=240
The zeros of n(n+1)=240 are n=15,n=-16