What are the solutions to the equation n(n+1)+3=45 ?

The solutions to the equation n(n+1)+3=45 are n=6,n=-7

Find the zeros of n(n+1)+3=45

The zeros of n(n+1)+3=45 are n=6,n=-7

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n (n + 1) + 3 = 45

n = 6, n = - 7

Solution steps

n (n + 1) + 3 = 45

Expand

n (n + 1) + 3 : n^{2} + n + 3

n^{2} + n + 3 = 45

Move

45

to the left side

n^{2} + n - 42 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 42 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 42) = 13

n_{1, 2} = \frac{- 1 \pm 13}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- 1 + 13}{2 \cdot 1}, n_{2} = \frac{- 1 - 13}{2 \cdot 1}

n = \frac{- 1 + 13}{2 \cdot 1} : 6

n = \frac{- 1 - 13}{2 \cdot 1} : - 7

The solutions to the quadratic equation are:

n = 6, n = - 7

6 x^{2} + 3 = - 2 x

8 x^{2} - 4 x + 6 = 0

4 x^{2} + 12 x - 27 = 13

z^{2} + 24 z + 140 = 0

98 x^{2} - 2 = 0

What are the solutions to the equation n(n+1)+3=45 ?
The solutions to the equation n(n+1)+3=45 are n=6,n=-7
Find the zeros of n(n+1)+3=45
The zeros of n(n+1)+3=45 are n=6,n=-7