What are the solutions to the equation 8b^2+11b-11=6 ?

The solutions to the equation 8b^2+11b-11=6 are b=(-11+sqrt(665))/(16),b=(-11-sqrt(665))/(16)

Find the zeros of 8b^2+11b-11=6

The zeros of 8b^2+11b-11=6 are b=(-11+sqrt(665))/(16),b=(-11-sqrt(665))/(16)

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8 b^{2} + 11 b - 11 = 6

b = \frac{- 11 + 665}{16}, b = \frac{- 11 - 665}{16}

Decimal Notation

b = 0.92422 \dots, b = - 2.29922 \dots

Solution steps

8 b^{2} + 11 b - 11 = 6

Move

6

to the left side

8 b^{2} + 11 b - 17 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 8 ( - 17 )}{2 \cdot 8}

1 1^{2} - 4 \cdot 8 (- 17) = 665

b_{1, 2} = \frac{- 11 \pm 665}{2 \cdot 8}

Separate the solutions

b_{1} = \frac{- 11 + 665}{2 \cdot 8}, b_{2} = \frac{- 11 - 665}{2 \cdot 8}

b = \frac{- 11 + 665}{2 \cdot 8} : \frac{- 11 + 665}{16}

b = \frac{- 11 - 665}{2 \cdot 8} : \frac{- 11 - 665}{16}

The solutions to the quadratic equation are:

b = \frac{- 11 + 665}{16}, b = \frac{- 11 - 665}{16}

3 x^{2} + 7 x - 20 = 0

2 m^{2} + 10 = 0

7 x + 7

\frac{5 x + 1}{3} < 7

- 1 - t \geq 1

What are the solutions to the equation 8b^2+11b-11=6 ?
The solutions to the equation 8b^2+11b-11=6 are b=(-11+sqrt(665))/(16),b=(-11-sqrt(665))/(16)
Find the zeros of 8b^2+11b-11=6
The zeros of 8b^2+11b-11=6 are b=(-11+sqrt(665))/(16),b=(-11-sqrt(665))/(16)