What are the solutions to the equation n(n+1)=0.3n(n)+300 ?

The solutions to the equation n(n+1)=0.3n(n)+300 are n=20,n=-150/7

Find the zeros of n(n+1)=0.3n(n)+300

The zeros of n(n+1)=0.3n(n)+300 are n=20,n=-150/7

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n (n + 1) = 0.3 n (n) + 300

n = 20, n = - \frac{150}{7}

Decimal Notation

n = 20, n = - 21.42857 \dots

Solution steps

n (n + 1) = 0.3 n (n) + 300

Multiply both sides by

10

n (n + 1) \cdot 10 = 3 n^{2} + 3000

Expand

n (n + 1) \cdot 10 : 10 n^{2} + 10 n

10 n^{2} + 10 n = 3 n^{2} + 3000

Move

3000

to the left side

10 n^{2} + 10 n - 3000 = 3 n^{2}

Move

3 n^{2}

to the left side

7 n^{2} + 10 n - 3000 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 7 ( - 3000 )}{2 \cdot 7}

1 0^{2} - 4 \cdot 7 (- 3000) = 290

n_{1, 2} = \frac{- 10 \pm 290}{2 \cdot 7}

Separate the solutions

n_{1} = \frac{- 10 + 290}{2 \cdot 7}, n_{2} = \frac{- 10 - 290}{2 \cdot 7}

n = \frac{- 10 + 290}{2 \cdot 7} : 20

n = \frac{- 10 - 290}{2 \cdot 7} : - \frac{150}{7}

The solutions to the quadratic equation are:

n = 20, n = - \frac{150}{7}

12 x^{2} + 29 x + 14 = 0

x, (x + 3)^{2} + (y + 5)^{2} + (z - 1)^{2} = 4

x (x + 2) = 2

y^{2} - 2 y - 3 = y - 3

x^{2} + 3 x - 36 = 0

What are the solutions to the equation n(n+1)=0.3n(n)+300 ?
The solutions to the equation n(n+1)=0.3n(n)+300 are n=20,n=-150/7
Find the zeros of n(n+1)=0.3n(n)+300
The zeros of n(n+1)=0.3n(n)+300 are n=20,n=-150/7