What are the solutions to the equation 78= 1/2 (n^2-n) ?

The solutions to the equation 78= 1/2 (n^2-n) are n=13,n=-12

Find the zeros of 78= 1/2 (n^2-n)

The zeros of 78= 1/2 (n^2-n) are n=13,n=-12

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78 = \frac{1}{2} (n^{2} - n)

n = 13, n = - 12

Solution steps

78 = \frac{1}{2} (n^{2} - n)

Multiply both sides by

2

156 = n^{2} - n

Switch sides

n^{2} - n = 156

Move

156

to the left side

n^{2} - n - 156 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 156 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 156) = 25

n_{1, 2} = \frac{- ( - 1 ) \pm 25}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- ( - 1 ) + 25}{2 \cdot 1}, n_{2} = \frac{- ( - 1 ) - 25}{2 \cdot 1}

n = \frac{- ( - 1 ) + 25}{2 \cdot 1} : 13

n = \frac{- ( - 1 ) - 25}{2 \cdot 1} : - 12

The solutions to the quadratic equation are:

n = 13, n = - 12

(1 0^{1})^{3}

x^{3} - 27 x < 0

\frac{1}{2} (1)

5 x (x + 2) + 6 = 3

x^{2} + 5 x + \frac{10}{4}

What are the solutions to the equation 78= 1/2 (n^2-n) ?
The solutions to the equation 78= 1/2 (n^2-n) are n=13,n=-12
Find the zeros of 78= 1/2 (n^2-n)
The zeros of 78= 1/2 (n^2-n) are n=13,n=-12