What are the solutions to the equation n^2-4n+9=0 ?

The solutions to the equation n^2-4n+9=0 are n=2+sqrt(5)i,n=2-sqrt(5)i

Find the zeros of n^2-4n+9=0

The zeros of n^2-4n+9=0 are n=2+sqrt(5)i,n=2-sqrt(5)i

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n^{2} - 4 n + 9 = 0

n = 2 + 5 i, n = 2 - 5 i

Solution steps

n^{2} - 4 n + 9 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot 9}{2 \cdot 1}

Simplify

(- 4)^{2} - 4 \cdot 1 \cdot 9 : 25 i

n_{1, 2} = \frac{- ( - 4 ) \pm 2 5 i}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- ( - 4 ) + 2 5 i}{2 \cdot 1}, n_{2} = \frac{- ( - 4 ) - 2 5 i}{2 \cdot 1}

n = \frac{- ( - 4 ) + 2 5 i}{2 \cdot 1} : 2 + 5 i

n = \frac{- ( - 4 ) - 2 5 i}{2 \cdot 1} : 2 - 5 i

The solutions to the quadratic equation are:

n = 2 + 5 i, n = 2 - 5 i

2 n^{2} - 8 n + 8

32 + 8

\frac{x}{3} - 4 = 2

\frac{10}{- 2}

\frac{8}{x + 4} \leq 2

What are the solutions to the equation n^2-4n+9=0 ?
The solutions to the equation n^2-4n+9=0 are n=2+sqrt(5)i,n=2-sqrt(5)i
Find the zeros of n^2-4n+9=0
The zeros of n^2-4n+9=0 are n=2+sqrt(5)i,n=2-sqrt(5)i