What are the solutions to the equation 56b^2+112b-102=3+7b^2 ?

The solutions to the equation 56b^2+112b-102=3+7b^2 are b= 5/7 ,b=-3

Find the zeros of 56b^2+112b-102=3+7b^2

The zeros of 56b^2+112b-102=3+7b^2 are b= 5/7 ,b=-3

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56 b^{2} + 112 b - 102 = 3 + 7 b^{2}

b = \frac{5}{7}, b = - 3

Decimal Notation

b = 0.71428 \dots, b = - 3

Solution steps

56 b^{2} + 112 b - 102 = 3 + 7 b^{2}

Move

7 b^{2}

to the left side

49 b^{2} + 112 b - 102 = 3

Move

3

to the left side

49 b^{2} + 112 b - 105 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- 112 \pm 11 2 ^{2} - 4 \cdot 49 ( - 105 )}{2 \cdot 49}

11 2^{2} - 4 \cdot 49 (- 105) = 182

b_{1, 2} = \frac{- 112 \pm 182}{2 \cdot 49}

Separate the solutions

b_{1} = \frac{- 112 + 182}{2 \cdot 49}, b_{2} = \frac{- 112 - 182}{2 \cdot 49}

b = \frac{- 112 + 182}{2 \cdot 49} : \frac{5}{7}

b = \frac{- 112 - 182}{2 \cdot 49} : - 3

The solutions to the quadratic equation are:

b = \frac{5}{7}, b = - 3

x, \frac{x ^{2}}{25} + \frac{y ^{2}}{16} = 0

8 x^{2} + 8 x + 8 = 0

x^{2} - 4 = - 2 x - 1

0 = 62181787.5 x - 458590682.8 x^{2} - 369609

4 x^{2} + 8 x - 12

What are the solutions to the equation 56b^2+112b-102=3+7b^2 ?
The solutions to the equation 56b^2+112b-102=3+7b^2 are b= 5/7 ,b=-3
Find the zeros of 56b^2+112b-102=3+7b^2
The zeros of 56b^2+112b-102=3+7b^2 are b= 5/7 ,b=-3