What are the solutions to the equation -8z=12+z^2 ?

The solutions to the equation -8z=12+z^2 are z=-2,z=-6

Find the zeros of -8z=12+z^2

The zeros of -8z=12+z^2 are z=-2,z=-6

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- 8 z = 12 + z^{2}

z = - 2, z = - 6

Solution steps

- 8 z = 12 + z^{2}

Switch sides

12 + z^{2} = - 8 z

Move

8 z

to the left side

12 + z^{2} + 8 z = 0

Write in the standard form

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

z^{2} + 8 z + 12 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 1 \cdot 12}{2 \cdot 1}

8^{2} - 4 \cdot 1 \cdot 12 = 4

z_{1, 2} = \frac{- 8 \pm 4}{2 \cdot 1}

Separate the solutions

z_{1} = \frac{- 8 + 4}{2 \cdot 1}, z_{2} = \frac{- 8 - 4}{2 \cdot 1}

z = \frac{- 8 + 4}{2 \cdot 1} : - 2

z = \frac{- 8 - 4}{2 \cdot 1} : - 6

The solutions to the quadratic equation are:

z = - 2, z = - 6

(x - 1)^{2} - (x - 1) = 42

x^{2} + 11 x - 17 = 0

5 x^{2} - 2 = - 9 x

y^{2} + 3 y = 2

(1 - x)^{2} = 9

What are the solutions to the equation -8z=12+z^2 ?
The solutions to the equation -8z=12+z^2 are z=-2,z=-6
Find the zeros of -8z=12+z^2
The zeros of -8z=12+z^2 are z=-2,z=-6