What are the solutions to the equation 7b^2=-9b+5 ?

The solutions to the equation 7b^2=-9b+5 are b=(-9+sqrt(221))/(14),b=(-9-sqrt(221))/(14)

Find the zeros of 7b^2=-9b+5

The zeros of 7b^2=-9b+5 are b=(-9+sqrt(221))/(14),b=(-9-sqrt(221))/(14)

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7 b^{2} = - 9 b + 5

b = \frac{- 9 + 221}{14}, b = \frac{- 9 - 221}{14}

Solution steps

7 b^{2} = - 9 b + 5

Move

5

to the left side

7 b^{2} - 5 = - 9 b

Move

9 b

to the left side

7 b^{2} - 5 + 9 b = 0

Write in the standard form

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

7 b^{2} + 9 b - 5 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 7 ( - 5 )}{2 \cdot 7}

9^{2} - 4 \cdot 7 (- 5) = 221

b_{1, 2} = \frac{- 9 \pm 221}{2 \cdot 7}

Separate the solutions

b_{1} = \frac{- 9 + 221}{2 \cdot 7}, b_{2} = \frac{- 9 - 221}{2 \cdot 7}

b = \frac{- 9 + 221}{2 \cdot 7} : \frac{- 9 + 221}{14}

b = \frac{- 9 - 221}{2 \cdot 7} : \frac{- 9 - 221}{14}

The solutions to the quadratic equation are:

b = \frac{- 9 + 221}{14}, b = \frac{- 9 - 221}{14}

9 = 39 - 4.9 (x)^{2}

(1 + \frac{0.05}{12})^{12} - 1

- 2 x = 38

x^{3} - 3 x^{2} - 9 x - 5

- \frac{1}{4} x^{2} + x + 1 = 0

What are the solutions to the equation 7b^2=-9b+5 ?
The solutions to the equation 7b^2=-9b+5 are b=(-9+sqrt(221))/(14),b=(-9-sqrt(221))/(14)
Find the zeros of 7b^2=-9b+5
The zeros of 7b^2=-9b+5 are b=(-9+sqrt(221))/(14),b=(-9-sqrt(221))/(14)