What are the solutions to the equation 6=(n(n-1))/2 ?

The solutions to the equation 6=(n(n-1))/2 are n=4,n=-3

Find the zeros of 6=(n(n-1))/2

The zeros of 6=(n(n-1))/2 are n=4,n=-3

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6 = \frac{n ( n - 1 )}{2}

n = 4, n = - 3

Solution steps

6 = \frac{n ( n - 1 )}{2}

Multiply both sides by

2

12 = n (n - 1)

Expand

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

12 = n^{2} - n

Switch sides

n^{2} - n = 12

Move

12

to the left side

n^{2} - n - 12 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 12 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 12) = 7

n_{1, 2} = \frac{- ( - 1 ) \pm 7}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- ( - 1 ) + 7}{2 \cdot 1}, n_{2} = \frac{- ( - 1 ) - 7}{2 \cdot 1}

n = \frac{- ( - 1 ) + 7}{2 \cdot 1} : 4

n = \frac{- ( - 1 ) - 7}{2 \cdot 1} : - 3

The solutions to the quadratic equation are:

n = 4, n = - 3

3^{3} (3 x)^{2} = 3^{5}

c, c^{2} = h^{2} - c^{2}

81 y^{2} - k = (\frac{9 y + 2}{3}) (\frac{9 y - 2}{3})

5 t^{2} - 20 t - 25 = 0

x, f (x + h) - f = 3 x - 4 x^{2}

What are the solutions to the equation 6=(n(n-1))/2 ?
The solutions to the equation 6=(n(n-1))/2 are n=4,n=-3
Find the zeros of 6=(n(n-1))/2
The zeros of 6=(n(n-1))/2 are n=4,n=-3