What are the solutions to the equation 7b^2+11b=-6 ?

The solutions to the equation 7b^2+11b=-6 are b=-11/14+i(sqrt(47))/(14),b=-11/14-i(sqrt(47))/(14)

Find the zeros of 7b^2+11b=-6

The zeros of 7b^2+11b=-6 are b=-11/14+i(sqrt(47))/(14),b=-11/14-i(sqrt(47))/(14)

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

b = - \frac{11}{14} + i \frac{47}{14}, b = - \frac{11}{14} - i \frac{47}{14}

Solution steps

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

Move

6

to the left side

7 b^{2} + 11 b + 6 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 7 \cdot 6}{2 \cdot 7}

Simplify

1 1^{2} - 4 \cdot 7 \cdot 6 : 47 i

b_{1, 2} = \frac{- 11 \pm 47 i}{2 \cdot 7}

Separate the solutions

b_{1} = \frac{- 11 + 47 i}{2 \cdot 7}, b_{2} = \frac{- 11 - 47 i}{2 \cdot 7}

b = \frac{- 11 + 47 i}{2 \cdot 7} : - \frac{11}{14} + i \frac{47}{14}

b = \frac{- 11 - 47 i}{2 \cdot 7} : - \frac{11}{14} - i \frac{47}{14}

The solutions to the quadratic equation are:

b = - \frac{11}{14} + i \frac{47}{14}, b = - \frac{11}{14} - i \frac{47}{14}

x^{2} - 12 x - 4 = 0

\frac{- x}{4} = \frac{- x}{3} + \frac{1}{4}

\frac{4 15}{2 3}

- 5 = - 3 (2) + b

3 x^{2} + x y - 2 y^{2}

What are the solutions to the equation 7b^2+11b=-6 ?
The solutions to the equation 7b^2+11b=-6 are b=-11/14+i(sqrt(47))/(14),b=-11/14-i(sqrt(47))/(14)
Find the zeros of 7b^2+11b=-6
The zeros of 7b^2+11b=-6 are b=-11/14+i(sqrt(47))/(14),b=-11/14-i(sqrt(47))/(14)