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

The solutions to the equation solvefor n,d=(n(n-3))/2 are n=(3+sqrt(9+8d))/2 ,n=(3-sqrt(9+8d))/2

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

The zeros of solvefor n,d=(n(n-3))/2 are n=(3+sqrt(9+8d))/2 ,n=(3-sqrt(9+8d))/2

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solve for

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

n = \frac{3 + 9 + 8 d}{2}, n = \frac{3 - 9 + 8 d}{2}

Solution steps

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

Multiply both sides by

2

2 d = n (n - 3)

Expand

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

2 d = n^{2} - 3 n

Switch sides

n^{2} - 3 n = 2 d

Move

2 d

to the left side

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

Solve with the quadratic formula

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

Simplify

(- 3)^{2} - 4 \cdot 1 \cdot (- 2 d) : 9 + 8 d

n_{1, 2} = \frac{- ( - 3 ) \pm 9 + 8 d}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- ( - 3 ) + 9 + 8 d}{2 \cdot 1}, n_{2} = \frac{- ( - 3 ) - 9 + 8 d}{2 \cdot 1}

n = \frac{- ( - 3 ) + 9 + 8 d}{2 \cdot 1} : \frac{3 + 9 + 8 d}{2}

n = \frac{- ( - 3 ) - 9 + 8 d}{2 \cdot 1} : \frac{3 - 9 + 8 d}{2}

The solutions to the quadratic equation are:

n = \frac{3 + 9 + 8 d}{2}, n = \frac{3 - 9 + 8 d}{2}

2 b^{2} + 15 = 16 b

x^{2} - x - 6 = 6

x, f^{- 1} (x) = x^{2} - 4

x^{2} - 18 x + 81 = 0

- 6 x^{2} + 18 x - 12 = 0

What are the solutions to the equation solvefor n,d=(n(n-3))/2 ?
The solutions to the equation solvefor n,d=(n(n-3))/2 are n=(3+sqrt(9+8d))/2 ,n=(3-sqrt(9+8d))/2
Find the zeros of solvefor n,d=(n(n-3))/2
The zeros of solvefor n,d=(n(n-3))/2 are n=(3+sqrt(9+8d))/2 ,n=(3-sqrt(9+8d))/2