What are the solutions to the equation 9z^2+29=36z ?

The solutions to the equation 9z^2+29=36z are z=(6+sqrt(7))/3 ,z=(6-sqrt(7))/3

Find the zeros of 9z^2+29=36z

The zeros of 9z^2+29=36z are z=(6+sqrt(7))/3 ,z=(6-sqrt(7))/3

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9 z^{2} + 29 = 36 z

z = \frac{6 + 7}{3}, z = \frac{6 - 7}{3}

Decimal Notation

z = 2.88191 \dots, z = 1.11808 \dots

Solution steps

9 z^{2} + 29 = 36 z

Move

36 z

to the left side

9 z^{2} + 29 - 36 z = 0

Write in the standard form

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

9 z^{2} - 36 z + 29 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- ( - 36 ) \pm ( - 36 ) ^{2} - 4 \cdot 9 \cdot 29}{2 \cdot 9}

(- 36)^{2} - 4 \cdot 9 \cdot 29 = 67

z_{1, 2} = \frac{- ( - 36 ) \pm 6 7}{2 \cdot 9}

Separate the solutions

z_{1} = \frac{- ( - 36 ) + 6 7}{2 \cdot 9}, z_{2} = \frac{- ( - 36 ) - 6 7}{2 \cdot 9}

z = \frac{- ( - 36 ) + 6 7}{2 \cdot 9} : \frac{6 + 7}{3}

z = \frac{- ( - 36 ) - 6 7}{2 \cdot 9} : \frac{6 - 7}{3}

The solutions to the quadratic equation are:

z = \frac{6 + 7}{3}, z = \frac{6 - 7}{3}

a^{5} \cdot a^{- 3}

(X + 5) (X - 1) (X - 2) > 0

ln (16)

\frac{x}{2} + 3 x (x - 1) = 38

7 - \frac{x}{3} > 8

What are the solutions to the equation 9z^2+29=36z ?
The solutions to the equation 9z^2+29=36z are z=(6+sqrt(7))/3 ,z=(6-sqrt(7))/3
Find the zeros of 9z^2+29=36z
The zeros of 9z^2+29=36z are z=(6+sqrt(7))/3 ,z=(6-sqrt(7))/3