What are the solutions to the equation 3(4b^2-7)=-2b-8 ?

The solutions to the equation 3(4b^2-7)=-2b-8 are b=(-1+sqrt(157))/(12),b=-(1+sqrt(157))/(12)

Find the zeros of 3(4b^2-7)=-2b-8

The zeros of 3(4b^2-7)=-2b-8 are b=(-1+sqrt(157))/(12),b=-(1+sqrt(157))/(12)

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3 (4 b^{2} - 7) = - 2 b - 8

b = \frac{- 1 + 157}{12}, b = - \frac{1 + 157}{12}

Solution steps

3 (4 b^{2} - 7) = - 2 b - 8

Expand

3 (4 b^{2} - 7) : 12 b^{2} - 21

12 b^{2} - 21 = - 2 b - 8

Move

8

to the left side

12 b^{2} - 13 = - 2 b

Move

2 b

to the left side

12 b^{2} - 13 + 2 b = 0

Write in the standard form

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

12 b^{2} + 2 b - 13 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 12 ( - 13 )}{2 \cdot 12}

2^{2} - 4 \cdot 12 (- 13) = 2157

b_{1, 2} = \frac{- 2 \pm 2 157}{2 \cdot 12}

Separate the solutions

b_{1} = \frac{- 2 + 2 157}{2 \cdot 12}, b_{2} = \frac{- 2 - 2 157}{2 \cdot 12}

b = \frac{- 2 + 2 157}{2 \cdot 12} : \frac{- 1 + 157}{12}

b = \frac{- 2 - 2 157}{2 \cdot 12} : - \frac{1 + 157}{12}

The solutions to the quadratic equation are:

b = \frac{- 1 + 157}{12}, b = - \frac{1 + 157}{12}

4 x^{2} + 20 x

x - 7 = (x - 1)^{2}

x, 25 x^{2} + k x + 4 = 0

4 x^{2} + 20 = 10 x + 3 x^{2} - 4

3 q^{2} - 147 = 0

What are the solutions to the equation 3(4b^2-7)=-2b-8 ?
The solutions to the equation 3(4b^2-7)=-2b-8 are b=(-1+sqrt(157))/(12),b=-(1+sqrt(157))/(12)
Find the zeros of 3(4b^2-7)=-2b-8
The zeros of 3(4b^2-7)=-2b-8 are b=(-1+sqrt(157))/(12),b=-(1+sqrt(157))/(12)