What are the solutions to the equation 3(6m-17)m=5 ?

The solutions to the equation 3(6m-17)m=5 are m=(17+sqrt(329))/(12),m=(17-sqrt(329))/(12)

Find the zeros of 3(6m-17)m=5

The zeros of 3(6m-17)m=5 are m=(17+sqrt(329))/(12),m=(17-sqrt(329))/(12)

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3 (6 m - 17) m = 5

m = \frac{17 + 329}{12}, m = \frac{17 - 329}{12}

Decimal Notation

m = 2.92819 \dots, m = - 0.09486 \dots

Solution steps

3 (6 m - 17) m = 5

Expand

3 (6 m - 17) m : 18 m^{2} - 51 m

18 m^{2} - 51 m = 5

Move

5

to the left side

18 m^{2} - 51 m - 5 = 0

Solve with the quadratic formula

m_{1, 2} = \frac{- ( - 51 ) \pm ( - 51 ) ^{2} - 4 \cdot 18 ( - 5 )}{2 \cdot 18}

(- 51)^{2} - 4 \cdot 18 (- 5) = 3329

m_{1, 2} = \frac{- ( - 51 ) \pm 3 329}{2 \cdot 18}

Separate the solutions

m_{1} = \frac{- ( - 51 ) + 3 329}{2 \cdot 18}, m_{2} = \frac{- ( - 51 ) - 3 329}{2 \cdot 18}

m = \frac{- ( - 51 ) + 3 329}{2 \cdot 18} : \frac{17 + 329}{12}

m = \frac{- ( - 51 ) - 3 329}{2 \cdot 18} : \frac{17 - 329}{12}

The solutions to the quadratic equation are:

m = \frac{17 + 329}{12}, m = \frac{17 - 329}{12}

4 u^{2} - 15 u - 4 = 0

2 x^{2} + 12 x + 21 = 0

- 0.012 x^{2} - 8 x + 430 = 0

z^{2} + 4 z + 16 = 0

x^{2} - 16 x + 64 = 12

What are the solutions to the equation 3(6m-17)m=5 ?
The solutions to the equation 3(6m-17)m=5 are m=(17+sqrt(329))/(12),m=(17-sqrt(329))/(12)
Find the zeros of 3(6m-17)m=5
The zeros of 3(6m-17)m=5 are m=(17+sqrt(329))/(12),m=(17-sqrt(329))/(12)