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

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

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

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

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

n = 4, n = 1

Solution steps

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

Multiply both sides by

2

2 n - 4 = n (n - 3)

Expand

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

2 n - 4 = n^{2} - 3 n

Switch sides

n^{2} - 3 n = 2 n - 4

Move

4

to the left side

n^{2} - 3 n + 4 = 2 n

Move

2 n

to the left side

n^{2} - 5 n + 4 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot 4}{2 \cdot 1}

(- 5)^{2} - 4 \cdot 1 \cdot 4 = 3

n_{1, 2} = \frac{- ( - 5 ) \pm 3}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- ( - 5 ) + 3}{2 \cdot 1}, n_{2} = \frac{- ( - 5 ) - 3}{2 \cdot 1}

n = \frac{- ( - 5 ) + 3}{2 \cdot 1} : 4

n = \frac{- ( - 5 ) - 3}{2 \cdot 1} : 1

The solutions to the quadratic equation are:

n = 4, n = 1

\frac{π 6 \cdot 4 5 ^{\circ}}{18 0 ^{\circ}}

x^{2} - 8 x - 33 = 0

(\frac{5}{2})^{2} - 5 (\frac{5}{2}) + 6

0. 5^{9}

4 ∣ x + 8 ∣ + 9 > 7

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