What are the solutions to the equation (n+4)+n(n+2)+n=0 ?

The solutions to the equation (n+4)+n(n+2)+n=0 are n=-2

Find the zeros of (n+4)+n(n+2)+n=0

The zeros of (n+4)+n(n+2)+n=0 are n=-2

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(n + 4) + n (n + 2) + n = 0

n = - 2

Solution steps

(n + 4) + n (n + 2) + n = 0

Expand

(n + 4) + n (n + 2) + n : n^{2} + 4 n + 4

n^{2} + 4 n + 4 = 0

Solve with the quadratic formula

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

4^{2} - 4 \cdot 1 \cdot 4 = 0

n_{1, 2} = \frac{- 4 \pm 0}{2 \cdot 1}

n = \frac{- 4}{2 \cdot 1}

\frac{- 4}{2 \cdot 1} = - 2

n = - 2

The solution to the quadratic equation is:

n = - 2

h^{2} = 32 - 4 h

4 r^{2} = 25

- \frac{1}{2} x^{2} - 2 x + 6 = 0

(130 π \cdot 1 0^{3} - w)^{2} = 3 \cdot 150 π \cdot 1 0^{3}

4 x^{2} + 7 = 0

What are the solutions to the equation (n+4)+n(n+2)+n=0 ?
The solutions to the equation (n+4)+n(n+2)+n=0 are n=-2
Find the zeros of (n+4)+n(n+2)+n=0
The zeros of (n+4)+n(n+2)+n=0 are n=-2