What are the solutions to the equation 9z^2+34=36z ?

The solutions to the equation 9z^2+34=36z are z=(6+sqrt(2))/3 ,z=(6-sqrt(2))/3

Find the zeros of 9z^2+34=36z

The zeros of 9z^2+34=36z are z=(6+sqrt(2))/3 ,z=(6-sqrt(2))/3

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9 z^{2} + 34 = 36 z

z = \frac{6 + 2}{3}, z = \frac{6 - 2}{3}

Decimal Notation

z = 2.47140 \dots, z = 1.52859 \dots

Solution steps

9 z^{2} + 34 = 36 z

Move

36 z

to the left side

9 z^{2} + 34 - 36 z = 0

Write in the standard form

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

9 z^{2} - 36 z + 34 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- ( - 36 ) \pm ( - 36 ) ^{2} - 4 \cdot 9 \cdot 34}{2 \cdot 9}

(- 36)^{2} - 4 \cdot 9 \cdot 34 = 62

z_{1, 2} = \frac{- ( - 36 ) \pm 6 2}{2 \cdot 9}

Separate the solutions

z_{1} = \frac{- ( - 36 ) + 6 2}{2 \cdot 9}, z_{2} = \frac{- ( - 36 ) - 6 2}{2 \cdot 9}

z = \frac{- ( - 36 ) + 6 2}{2 \cdot 9} : \frac{6 + 2}{3}

z = \frac{- ( - 36 ) - 6 2}{2 \cdot 9} : \frac{6 - 2}{3}

The solutions to the quadratic equation are:

z = \frac{6 + 2}{3}, z = \frac{6 - 2}{3}

(x + y)^{4} - 1

1 - \frac{8}{3}

3 x^{2} + 6 x + 7 = 0

4 x^{4} - 25 x^{2} + 36

\frac{x}{8} = \frac{3}{5}

What are the solutions to the equation 9z^2+34=36z ?
The solutions to the equation 9z^2+34=36z are z=(6+sqrt(2))/3 ,z=(6-sqrt(2))/3
Find the zeros of 9z^2+34=36z
The zeros of 9z^2+34=36z are z=(6+sqrt(2))/3 ,z=(6-sqrt(2))/3