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

The solutions to the equation 2+z=6z^2 are z= 2/3 ,z=-1/2

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

The zeros of 2+z=6z^2 are z= 2/3 ,z=-1/2

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2 + z = 6 z^{2}

z = \frac{2}{3}, z = - \frac{1}{2}

Solution steps

2 + z = 6 z^{2}

Switch sides

6 z^{2} = 2 + z

Move

z

to the left side

6 z^{2} - z = 2

Move

2

to the left side

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

Solve with the quadratic formula

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

(- 1)^{2} - 4 \cdot 6 (- 2) = 7

z_{1, 2} = \frac{- ( - 1 ) \pm 7}{2 \cdot 6}

Separate the solutions

z_{1} = \frac{- ( - 1 ) + 7}{2 \cdot 6}, z_{2} = \frac{- ( - 1 ) - 7}{2 \cdot 6}

z = \frac{- ( - 1 ) + 7}{2 \cdot 6} : \frac{2}{3}

z = \frac{- ( - 1 ) - 7}{2 \cdot 6} : - \frac{1}{2}

The solutions to the quadratic equation are:

z = \frac{2}{3}, z = - \frac{1}{2}

- 17 = \frac{x}{5}

12 c - 20

3 x + 2 < 10

- \frac{x}{11} = \frac{1}{26}

(t^{2})^{2}

What are the solutions to the equation 2+z=6z^2 ?
The solutions to the equation 2+z=6z^2 are z= 2/3 ,z=-1/2
Find the zeros of 2+z=6z^2
The zeros of 2+z=6z^2 are z= 2/3 ,z=-1/2