What are the solutions to the equation b-3(b-3)(1-b)=42 ?

The solutions to the equation b-3(b-3)(1-b)=42 are b=(11+sqrt(517))/6 ,b=(11-sqrt(517))/6

Find the zeros of b-3(b-3)(1-b)=42

The zeros of b-3(b-3)(1-b)=42 are b=(11+sqrt(517))/6 ,b=(11-sqrt(517))/6

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b - 3 (b - 3) (1 - b) = 42

b = \frac{11 + 517}{6}, b = \frac{11 - 517}{6}

Solution steps

b - 3 (b - 3) (1 - b) = 42

Expand

b - 3 (b - 3) (1 - b) : - 11 b + 3 b^{2} + 9

- 11 b + 3 b^{2} + 9 = 42

Move

42

to the left side

3 b^{2} - 11 b - 33 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 3 ( - 33 )}{2 \cdot 3}

(- 11)^{2} - 4 \cdot 3 (- 33) = 517

b_{1, 2} = \frac{- ( - 11 ) \pm 517}{2 \cdot 3}

Separate the solutions

b_{1} = \frac{- ( - 11 ) + 517}{2 \cdot 3}, b_{2} = \frac{- ( - 11 ) - 517}{2 \cdot 3}

b = \frac{- ( - 11 ) + 517}{2 \cdot 3} : \frac{11 + 517}{6}

b = \frac{- ( - 11 ) - 517}{2 \cdot 3} : \frac{11 - 517}{6}

The solutions to the quadratic equation are:

b = \frac{11 + 517}{6}, b = \frac{11 - 517}{6}

- 3 x - 1 = - 8 x + 34

\frac{1}{72}

\frac{( \frac{4}{5} )}{( \frac{3}{4} )}

- 2 \cdot - 18

x^{2} x

What are the solutions to the equation b-3(b-3)(1-b)=42 ?
The solutions to the equation b-3(b-3)(1-b)=42 are b=(11+sqrt(517))/6 ,b=(11-sqrt(517))/6
Find the zeros of b-3(b-3)(1-b)=42
The zeros of b-3(b-3)(1-b)=42 are b=(11+sqrt(517))/6 ,b=(11-sqrt(517))/6