What are the solutions to the equation (n(n+1))/2 =100 ?

The solutions to the equation (n(n+1))/2 =100 are n=(-1+3sqrt(89))/2 ,n=(-1-3sqrt(89))/2

Find the zeros of (n(n+1))/2 =100

The zeros of (n(n+1))/2 =100 are n=(-1+3sqrt(89))/2 ,n=(-1-3sqrt(89))/2

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\frac{n ( n + 1 )}{2} = 100

n = \frac{- 1 + 3 89}{2}, n = \frac{- 1 - 3 89}{2}

Decimal Notation

n = 13.65097 \dots, n = - 14.65097 \dots

Solution steps

\frac{n ( n + 1 )}{2} = 100

Multiply both sides by

2

n (n + 1) = 200

Expand

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

n^{2} + n = 200

Move

200

to the left side

n^{2} + n - 200 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 200 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 200) = 389

n_{1, 2} = \frac{- 1 \pm 3 89}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- 1 + 3 89}{2 \cdot 1}, n_{2} = \frac{- 1 - 3 89}{2 \cdot 1}

n = \frac{- 1 + 3 89}{2 \cdot 1} : \frac{- 1 + 3 89}{2}

n = \frac{- 1 - 3 89}{2 \cdot 1} : \frac{- 1 - 3 89}{2}

The solutions to the quadratic equation are:

n = \frac{- 1 + 3 89}{2}, n = \frac{- 1 - 3 89}{2}

121 x^{2} + 100 = 0

- 3 t^{2} + 9 t = - 810

- 8 π r^{2} + 176 π r - 768 π = 0

1 + 2 x - 3 x^{2} = 0

2 z^{2} + 4 z + 2 = 0

What are the solutions to the equation (n(n+1))/2 =100 ?
The solutions to the equation (n(n+1))/2 =100 are n=(-1+3sqrt(89))/2 ,n=(-1-3sqrt(89))/2
Find the zeros of (n(n+1))/2 =100
The zeros of (n(n+1))/2 =100 are n=(-1+3sqrt(89))/2 ,n=(-1-3sqrt(89))/2