What are the solutions to the equation n^2+14n+48=0 ?

The solutions to the equation n^2+14n+48=0 are n=-6,n=-8

Find the zeros of n^2+14n+48=0

The zeros of n^2+14n+48=0 are n=-6,n=-8

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n^{2} + 14 n + 48 = 0

n = - 6, n = - 8

Solution steps

n^{2} + 14 n + 48 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 14 \pm 1 4 ^{2} - 4 \cdot 1 \cdot 48}{2 \cdot 1}

1 4^{2} - 4 \cdot 1 \cdot 48 = 2

n_{1, 2} = \frac{- 14 \pm 2}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- 14 + 2}{2 \cdot 1}, n_{2} = \frac{- 14 - 2}{2 \cdot 1}

n = \frac{- 14 + 2}{2 \cdot 1} : - 6

n = \frac{- 14 - 2}{2 \cdot 1} : - 8

The solutions to the quadratic equation are:

n = - 6, n = - 8

3 x^{2} + 12 x + 3

0 = - 3010 + 497 x - \frac{93}{5} x^{2}

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

(x - 8) (5 x - 6) = 0

2 x^{2} + 20 x = 20

What are the solutions to the equation n^2+14n+48=0 ?
The solutions to the equation n^2+14n+48=0 are n=-6,n=-8
Find the zeros of n^2+14n+48=0
The zeros of n^2+14n+48=0 are n=-6,n=-8