What are the solutions to the equation n^2-12n+27=0 ?

The solutions to the equation n^2-12n+27=0 are n=9,n=3

Find the zeros of n^2-12n+27=0

The zeros of n^2-12n+27=0 are n=9,n=3

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n^{2} - 12 n + 27 = 0

n = 9, n = 3

Solution steps

n^{2} - 12 n + 27 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 1 \cdot 27}{2 \cdot 1}

(- 12)^{2} - 4 \cdot 1 \cdot 27 = 6

n_{1, 2} = \frac{- ( - 12 ) \pm 6}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- ( - 12 ) + 6}{2 \cdot 1}, n_{2} = \frac{- ( - 12 ) - 6}{2 \cdot 1}

n = \frac{- ( - 12 ) + 6}{2 \cdot 1} : 9

n = \frac{- ( - 12 ) - 6}{2 \cdot 1} : 3

The solutions to the quadratic equation are:

n = 9, n = 3

\frac{2}{3} s + 12 = \frac{- 2}{3} s + 14

- \frac{7}{8} - \frac{1}{2}

\frac{7 x - 1}{6} - \frac{2 x - 1}{2} \leq \frac{1}{2}

\frac{12 - 7.35}{4.11}

12 x^{2} + 6 x

What are the solutions to the equation n^2-12n+27=0 ?
The solutions to the equation n^2-12n+27=0 are n=9,n=3
Find the zeros of n^2-12n+27=0
The zeros of n^2-12n+27=0 are n=9,n=3