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

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

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

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

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

n = 8, n = 6

Solution steps

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

Solve with the quadratic formula

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

(- 14)^{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} : 8

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

The solutions to the quadratic equation are:

n = 8, n = 6

9 ∣ m - 8 ∣ - 10 \leq 26

- \frac{5}{3} n + \frac{4}{3} + \frac{3}{4} = \frac{19}{4}

2 \leq \frac{x}{9} + 1

- 5 x^{2} - 24 x + 5 = 0

c + 23 = 10

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