What are the solutions to the equation 16(4-3m)=96(-m^2+1) ?

The solutions to the equation 16(4-3m)=96(-m^2+1) are m=-(-3+sqrt(57))/(12),m=(3+sqrt(57))/(12)

Find the zeros of 16(4-3m)=96(-m^2+1)

The zeros of 16(4-3m)=96(-m^2+1) are m=-(-3+sqrt(57))/(12),m=(3+sqrt(57))/(12)

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16 (4 - 3 m) = 96 (- m^{2} + 1)

m = - \frac{- 3 + 57}{12}, m = \frac{3 + 57}{12}

Solution steps

16 (4 - 3 m) = 96 (- m^{2} + 1)

Expand

16 (4 - 3 m) : 64 - 48 m

Expand

96 (- m^{2} + 1) : - 96 m^{2} + 96

64 - 48 m = - 96 m^{2} + 96

Switch sides

- 96 m^{2} + 96 = 64 - 48 m

Move

48 m

to the left side

- 96 m^{2} + 96 + 48 m = 64

Move

64

to the left side

- 96 m^{2} + 48 m + 32 = 0

Solve with the quadratic formula

m_{1, 2} = \frac{- 48 \pm 4 8 ^{2} - 4 ( - 96 ) \cdot 32}{2 ( - 96 )}

4 8^{2} - 4 (- 96) \cdot 32 = 1657

m_{1, 2} = \frac{- 48 \pm 16 57}{2 ( - 96 )}

Separate the solutions

m_{1} = \frac{- 48 + 16 57}{2 ( - 96 )}, m_{2} = \frac{- 48 - 16 57}{2 ( - 96 )}

m = \frac{- 48 + 16 57}{2 ( - 96 )} : - \frac{- 3 + 57}{12}

m = \frac{- 48 - 16 57}{2 ( - 96 )} : \frac{3 + 57}{12}

The solutions to the quadratic equation are:

m = - \frac{- 3 + 57}{12}, m = \frac{3 + 57}{12}

(5 x - 3) (x + 6) = 0

5 w^{2} + 13 w = - 15

x (5 x - 29) = 42

- 8 x^{2} + 13 = - 1139

11 x^{2} + 30 x - 1 = 0

What are the solutions to the equation 16(4-3m)=96(-m^2+1) ?
The solutions to the equation 16(4-3m)=96(-m^2+1) are m=-(-3+sqrt(57))/(12),m=(3+sqrt(57))/(12)
Find the zeros of 16(4-3m)=96(-m^2+1)
The zeros of 16(4-3m)=96(-m^2+1) are m=-(-3+sqrt(57))/(12),m=(3+sqrt(57))/(12)