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

The solutions to the equation (n(n-3))/2 =2n are n=7,n=0

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

The zeros of (n(n-3))/2 =2n are n=7,n=0

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

n = 7, n = 0

Solution steps

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

Multiply both sides by

2

n (n - 3) = 4 n

Expand

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

n^{2} - 3 n = 4 n

Move

4 n

to the left side

n^{2} - 7 n = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

(- 7)^{2} - 4 \cdot 1 \cdot 0 = 7

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

Separate the solutions

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

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

n = \frac{- ( - 7 ) - 7}{2 \cdot 1} : 0

The solutions to the quadratic equation are:

n = 7, n = 0

\frac{n ( n - 3 )}{2} = 27

2 x^{2} - 10 = 64

x^{2} - 8 x + 45 = 0

x^{2} + 7 x - 18 = 0

81 x^{2} - 180 x + 100 = 100

What are the solutions to the equation (n(n-3))/2 =2n ?
The solutions to the equation (n(n-3))/2 =2n are n=7,n=0
Find the zeros of (n(n-3))/2 =2n
The zeros of (n(n-3))/2 =2n are n=7,n=0