What are the solutions to the equation z^2-17z+12=-5z ?

The solutions to the equation z^2-17z+12=-5z are z=2(3+sqrt(6)),z=2(3-sqrt(6))

Find the zeros of z^2-17z+12=-5z

The zeros of z^2-17z+12=-5z are z=2(3+sqrt(6)),z=2(3-sqrt(6))

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z^{2} - 17 z + 12 = - 5 z

z = 2 (3 + 6), z = 2 (3 - 6)

Decimal Notation

z = 10.89897 \dots, z = 1.10102 \dots

Solution steps

z^{2} - 17 z + 12 = - 5 z

Move

5 z

to the left side

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

Solve with the quadratic formula

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

(- 12)^{2} - 4 \cdot 1 \cdot 12 = 46

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

Separate the solutions

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

z = \frac{- ( - 12 ) + 4 6}{2 \cdot 1} : 2 (3 + 6)

z = \frac{- ( - 12 ) - 4 6}{2 \cdot 1} : 2 (3 - 6)

The solutions to the quadratic equation are:

z = 2 (3 + 6), z = 2 (3 - 6)

3 x^{2} = - x + 4

x, y = 4 - x^{2}

99 a^{2} + 55 a = 0

2 x^{2} - 6 x = 6 \times 2 - 8 x

(x - 3)^{2} = - 25

What are the solutions to the equation z^2-17z+12=-5z ?
The solutions to the equation z^2-17z+12=-5z are z=2(3+sqrt(6)),z=2(3-sqrt(6))
Find the zeros of z^2-17z+12=-5z
The zeros of z^2-17z+12=-5z are z=2(3+sqrt(6)),z=2(3-sqrt(6))