What are the solutions to the equation n^2+n=1640 ?

The solutions to the equation n^2+n=1640 are n=40,n=-41

Find the zeros of n^2+n=1640

The zeros of n^2+n=1640 are n=40,n=-41

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n^{2} + n = 1640

n = 40, n = - 41

Solution steps

n^{2} + n = 1640

Move

1640

to the left side

n^{2} + n - 1640 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 1640 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 1640) = 81

n_{1, 2} = \frac{- 1 \pm 81}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- 1 + 81}{2 \cdot 1}, n_{2} = \frac{- 1 - 81}{2 \cdot 1}

n = \frac{- 1 + 81}{2 \cdot 1} : 40

n = \frac{- 1 - 81}{2 \cdot 1} : - 41

The solutions to the quadratic equation are:

n = 40, n = - 41

x^{2} + 7 x - 9 = - 3 + 6 x

n^{2} - 3 n + 8 n = 4

100 x^{2} - 60 x + 9 = 0

(a + 4) (a - 2) = 55

2 x (x - 6) = 10

What are the solutions to the equation n^2+n=1640 ?
The solutions to the equation n^2+n=1640 are n=40,n=-41
Find the zeros of n^2+n=1640
The zeros of n^2+n=1640 are n=40,n=-41