What are the solutions to the equation 16z^2+59=64z ?

The solutions to the equation 16z^2+59=64z are z=(8+sqrt(5))/4 ,z=(8-sqrt(5))/4

Find the zeros of 16z^2+59=64z

The zeros of 16z^2+59=64z are z=(8+sqrt(5))/4 ,z=(8-sqrt(5))/4

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16 z^{2} + 59 = 64 z

z = \frac{8 + 5}{4}, z = \frac{8 - 5}{4}

Solution steps

16 z^{2} + 59 = 64 z

Move

64 z

to the left side

16 z^{2} + 59 - 64 z = 0

Write in the standard form

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

16 z^{2} - 64 z + 59 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- ( - 64 ) \pm ( - 64 ) ^{2} - 4 \cdot 16 \cdot 59}{2 \cdot 16}

(- 64)^{2} - 4 \cdot 16 \cdot 59 = 85

z_{1, 2} = \frac{- ( - 64 ) \pm 8 5}{2 \cdot 16}

Separate the solutions

z_{1} = \frac{- ( - 64 ) + 8 5}{2 \cdot 16}, z_{2} = \frac{- ( - 64 ) - 8 5}{2 \cdot 16}

z = \frac{- ( - 64 ) + 8 5}{2 \cdot 16} : \frac{8 + 5}{4}

z = \frac{- ( - 64 ) - 8 5}{2 \cdot 16} : \frac{8 - 5}{4}

The solutions to the quadratic equation are:

z = \frac{8 + 5}{4}, z = \frac{8 - 5}{4}

\frac{2035}{1728} + \frac{1}{125} + \frac{1}{216} + \frac{1}{343} + \frac{1}{512}

- \frac{3}{4} - \frac{1}{4}

7 m + 14 n - 28

103 - 12

\frac{2 - 3 x}{4} > 5

What are the solutions to the equation 16z^2+59=64z ?
The solutions to the equation 16z^2+59=64z are z=(8+sqrt(5))/4 ,z=(8-sqrt(5))/4
Find the zeros of 16z^2+59=64z
The zeros of 16z^2+59=64z are z=(8+sqrt(5))/4 ,z=(8-sqrt(5))/4