What are the solutions to the equation z^2-12z+99=-3 ?

The solutions to the equation z^2-12z+99=-3 are z=6+sqrt(66)i,z=6-sqrt(66)i

Find the zeros of z^2-12z+99=-3

The zeros of z^2-12z+99=-3 are z=6+sqrt(66)i,z=6-sqrt(66)i

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z^{2} - 12 z + 99 = - 3

z = 6 + 66 i, z = 6 - 66 i

Solution steps

z^{2} - 12 z + 99 = - 3

Move

3

to the left side

z^{2} - 12 z + 102 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 1 \cdot 102}{2 \cdot 1}

Simplify

(- 12)^{2} - 4 \cdot 1 \cdot 102 : 266 i

z_{1, 2} = \frac{- ( - 12 ) \pm 2 66 i}{2 \cdot 1}

Separate the solutions

z_{1} = \frac{- ( - 12 ) + 2 66 i}{2 \cdot 1}, z_{2} = \frac{- ( - 12 ) - 2 66 i}{2 \cdot 1}

z = \frac{- ( - 12 ) + 2 66 i}{2 \cdot 1} : 6 + 66 i

z = \frac{- ( - 12 ) - 2 66 i}{2 \cdot 1} : 6 - 66 i

The solutions to the quadratic equation are:

z = 6 + 66 i, z = 6 - 66 i

lo g_{13} (16)

q^{2} - p^{2} + 2 p - 1

x - 15 = 0

2 x^{3} + x^{2} + x - 1

- 2 x^{2} - 16 x - 28 = 0

What are the solutions to the equation z^2-12z+99=-3 ?
The solutions to the equation z^2-12z+99=-3 are z=6+sqrt(66)i,z=6-sqrt(66)i
Find the zeros of z^2-12z+99=-3
The zeros of z^2-12z+99=-3 are z=6+sqrt(66)i,z=6-sqrt(66)i