What are the solutions to the equation z^2-6z+29=5 ?

The solutions to the equation z^2-6z+29=5 are z=3+sqrt(15)i,z=3-sqrt(15)i

Find the zeros of z^2-6z+29=5

The zeros of z^2-6z+29=5 are z=3+sqrt(15)i,z=3-sqrt(15)i

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z^{2} - 6 z + 29 = 5

z = 3 + 15 i, z = 3 - 15 i

Solution steps

z^{2} - 6 z + 29 = 5

Move

5

to the left side

z^{2} - 6 z + 24 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot 24}{2 \cdot 1}

Simplify

(- 6)^{2} - 4 \cdot 1 \cdot 24 : 215 i

z_{1, 2} = \frac{- ( - 6 ) \pm 2 15 i}{2 \cdot 1}

Separate the solutions

z_{1} = \frac{- ( - 6 ) + 2 15 i}{2 \cdot 1}, z_{2} = \frac{- ( - 6 ) - 2 15 i}{2 \cdot 1}

z = \frac{- ( - 6 ) + 2 15 i}{2 \cdot 1} : 3 + 15 i

z = \frac{- ( - 6 ) - 2 15 i}{2 \cdot 1} : 3 - 15 i

The solutions to the quadratic equation are:

z = 3 + 15 i, z = 3 - 15 i

e^{2} = 1

x^{2} - 4 x^{2} = 0

3 x^{2} - 4 x - 7 = 0

1 = n - 1 n - 4 + 1 n^{2} - n - 12

- x^{2} = - 10

What are the solutions to the equation z^2-6z+29=5 ?
The solutions to the equation z^2-6z+29=5 are z=3+sqrt(15)i,z=3-sqrt(15)i
Find the zeros of z^2-6z+29=5
The zeros of z^2-6z+29=5 are z=3+sqrt(15)i,z=3-sqrt(15)i