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

The solutions to the equation 3n=(n(n-3))/2 are n=9,n=0

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

The zeros of 3n=(n(n-3))/2 are n=9,n=0

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

n = 9, n = 0

Solution steps

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

Multiply both sides by

2

6 n = n (n - 3)

Expand

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

6 n = n^{2} - 3 n

Switch sides

n^{2} - 3 n = 6 n

Move

6 n

to the left side

n^{2} - 9 n = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

(- 9)^{2} - 4 \cdot 1 \cdot 0 = 9

n_{1, 2} = \frac{- ( - 9 ) \pm 9}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- ( - 9 ) + 9}{2 \cdot 1}, n_{2} = \frac{- ( - 9 ) - 9}{2 \cdot 1}

n = \frac{- ( - 9 ) + 9}{2 \cdot 1} : 9

n = \frac{- ( - 9 ) - 9}{2 \cdot 1} : 0

The solutions to the quadratic equation are:

n = 9, n = 0

2 (\frac{21}{29}) (\frac{20}{29})

5 p^{2} + 19 p + 12

12 1^{- \frac{1}{2}}

\frac{\frac{1}{32} - \frac{1}{8}}{\frac{1}{4}}

- 10 x^{2} + 13 x + 3 = 0

What are the solutions to the equation 3n=(n(n-3))/2 ?
The solutions to the equation 3n=(n(n-3))/2 are n=9,n=0
Find the zeros of 3n=(n(n-3))/2
The zeros of 3n=(n(n-3))/2 are n=9,n=0