What are the solutions to the equation (n(n-3))/2 =65 ?

The solutions to the equation (n(n-3))/2 =65 are n=13,n=-10

Find the zeros of (n(n-3))/2 =65

The zeros of (n(n-3))/2 =65 are n=13,n=-10

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\frac{n ( n - 3 )}{2} = 65

n = 13, n = - 10

Solution steps

\frac{n ( n - 3 )}{2} = 65

Multiply both sides by

2

n (n - 3) = 130

Expand

n (n - 3) : n^{2} - 3 n

n^{2} - 3 n = 130

Move

130

to the left side

n^{2} - 3 n - 130 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 130 )}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot (- 130) = 23

n_{1, 2} = \frac{- ( - 3 ) \pm 23}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- ( - 3 ) + 23}{2 \cdot 1}, n_{2} = \frac{- ( - 3 ) - 23}{2 \cdot 1}

n = \frac{- ( - 3 ) + 23}{2 \cdot 1} : 13

n = \frac{- ( - 3 ) - 23}{2 \cdot 1} : - 10

The solutions to the quadratic equation are:

n = 13, n = - 10

1 < 3 x + 2 < 3

∣ 8 y - 9 ∣ \geq 12

4 x^{2} < 1

- 9 x + 5 < 17

\frac{3}{4} - \frac{1}{3}

What are the solutions to the equation (n(n-3))/2 =65 ?
The solutions to the equation (n(n-3))/2 =65 are n=13,n=-10
Find the zeros of (n(n-3))/2 =65
The zeros of (n(n-3))/2 =65 are n=13,n=-10