What are the solutions to the equation z^2=-12z+6 ?

The solutions to the equation z^2=-12z+6 are z=-6+sqrt(42),z=-6-sqrt(42)

Find the zeros of z^2=-12z+6

The zeros of z^2=-12z+6 are z=-6+sqrt(42),z=-6-sqrt(42)

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z^{2} = - 12 z + 6

z = - 6 + 42, z = - 6 - 42

Decimal Notation

z = 0.48074 \dots, z = - 12.48074 \dots

Solution steps

z^{2} = - 12 z + 6

Move

6

to the left side

z^{2} - 6 = - 12 z

Move

12 z

to the left side

z^{2} - 6 + 12 z = 0

Write in the standard form

a x^{2} + b x + c = 0

z^{2} + 12 z - 6 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 1 \cdot ( - 6 )}{2 \cdot 1}

1 2^{2} - 4 \cdot 1 \cdot (- 6) = 242

z_{1, 2} = \frac{- 12 \pm 2 42}{2 \cdot 1}

Separate the solutions

z_{1} = \frac{- 12 + 2 42}{2 \cdot 1}, z_{2} = \frac{- 12 - 2 42}{2 \cdot 1}

z = \frac{- 12 + 2 42}{2 \cdot 1} : - 6 + 42

z = \frac{- 12 - 2 42}{2 \cdot 1} : - 6 - 42

The solutions to the quadratic equation are:

z = - 6 + 42, z = - 6 - 42

x^{2} + 10 x = - 12

x, x^{2} + y^{2} - 3 x + y + 10 = 0

x, x^{2} - y^{2} = 1

- 2 x^{2} + 10 = 0

- \frac{7}{4} x^{2} + \frac{3}{2} x = 0

What are the solutions to the equation z^2=-12z+6 ?
The solutions to the equation z^2=-12z+6 are z=-6+sqrt(42),z=-6-sqrt(42)
Find the zeros of z^2=-12z+6
The zeros of z^2=-12z+6 are z=-6+sqrt(42),z=-6-sqrt(42)