What are the solutions to the equation z^2+32=12z ?

The solutions to the equation z^2+32=12z are z=8,z=4

Find the zeros of z^2+32=12z

The zeros of z^2+32=12z are z=8,z=4

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z^{2} + 32 = 12 z

z = 8, z = 4

Solution steps

z^{2} + 32 = 12 z

Move

12 z

to the left side

z^{2} + 32 - 12 z = 0

Write in the standard form

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

z^{2} - 12 z + 32 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 1 \cdot 32}{2 \cdot 1}

(- 12)^{2} - 4 \cdot 1 \cdot 32 = 4

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

Separate the solutions

z_{1} = \frac{- ( - 12 ) + 4}{2 \cdot 1}, z_{2} = \frac{- ( - 12 ) - 4}{2 \cdot 1}

z = \frac{- ( - 12 ) + 4}{2 \cdot 1} : 8

z = \frac{- ( - 12 ) - 4}{2 \cdot 1} : 4

The solutions to the quadratic equation are:

z = 8, z = 4

\frac{3}{4} \cdot 64

(1 + 3) (1 - 3)

x^{2} - 27 = - 6 x

m^{3} - 64 m - 3 m^{2} + 192

56 = 7 + 7 y

What are the solutions to the equation z^2+32=12z ?
The solutions to the equation z^2+32=12z are z=8,z=4
Find the zeros of z^2+32=12z
The zeros of z^2+32=12z are z=8,z=4