What are the solutions to the equation n^2-4n+29=0 ?

The solutions to the equation n^2-4n+29=0 are n=2+5i,n=2-5i

Find the zeros of n^2-4n+29=0

The zeros of n^2-4n+29=0 are n=2+5i,n=2-5i

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n^{2} - 4 n + 29 = 0

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

Solution steps

n^{2} - 4 n + 29 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot 29}{2 \cdot 1}

Simplify

(- 4)^{2} - 4 \cdot 1 \cdot 29 : 10 i

n_{1, 2} = \frac{- ( - 4 ) \pm 10 i}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- ( - 4 ) + 10 i}{2 \cdot 1}, n_{2} = \frac{- ( - 4 ) - 10 i}{2 \cdot 1}

n = \frac{- ( - 4 ) + 10 i}{2 \cdot 1} : 2 + 5 i

n = \frac{- ( - 4 ) - 10 i}{2 \cdot 1} : 2 - 5 i

The solutions to the quadratic equation are:

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

X^{2} + 2 X - 24 = 0

- 0.0625 x^{2} + 2 x + 50 = 0

4 x (x + 5) = 6 x - 4 x^{2} - 3

6 x^{2} + 5 x - 4 = 0

(x + 7) (x + 3) = 10 (x + 2) + 2

What are the solutions to the equation n^2-4n+29=0 ?
The solutions to the equation n^2-4n+29=0 are n=2+5i,n=2-5i
Find the zeros of n^2-4n+29=0
The zeros of n^2-4n+29=0 are n=2+5i,n=2-5i