What are the solutions to the equation 44=(n(n-3))/2 ?

The solutions to the equation 44=(n(n-3))/2 are n=11,n=-8

Find the zeros of 44=(n(n-3))/2

The zeros of 44=(n(n-3))/2 are n=11,n=-8

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

n = 11, n = - 8

Solution steps

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

Multiply both sides by

2

88 = n (n - 3)

Expand

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

88 = n^{2} - 3 n

Switch sides

n^{2} - 3 n = 88

Move

88

to the left side

n^{2} - 3 n - 88 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 88 )}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot (- 88) = 19

n_{1, 2} = \frac{- ( - 3 ) \pm 19}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- ( - 3 ) + 19}{2 \cdot 1}, n_{2} = \frac{- ( - 3 ) - 19}{2 \cdot 1}

n = \frac{- ( - 3 ) + 19}{2 \cdot 1} : 11

n = \frac{- ( - 3 ) - 19}{2 \cdot 1} : - 8

The solutions to the quadratic equation are:

n = 11, n = - 8

15 x^{2} - 34 x - 16 = 0

36 t^{2} - 36 t + 2 = 0

14 x^{2} - 1 = 0

p, I = 40 p - p^{2}

y, z = 9 - y^{2}

What are the solutions to the equation 44=(n(n-3))/2 ?
The solutions to the equation 44=(n(n-3))/2 are n=11,n=-8
Find the zeros of 44=(n(n-3))/2
The zeros of 44=(n(n-3))/2 are n=11,n=-8