What are the solutions to the equation 3b^2+1=-2b ?

The solutions to the equation 3b^2+1=-2b are b=-1/3+i(sqrt(2))/3 ,b=-1/3-i(sqrt(2))/3

Find the zeros of 3b^2+1=-2b

The zeros of 3b^2+1=-2b are b=-1/3+i(sqrt(2))/3 ,b=-1/3-i(sqrt(2))/3

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3 b^{2} + 1 = - 2 b

b = - \frac{1}{3} + i \frac{2}{3}, b = - \frac{1}{3} - i \frac{2}{3}

Solution steps

3 b^{2} + 1 = - 2 b

Move

2 b

to the left side

3 b^{2} + 1 + 2 b = 0

Write in the standard form

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

3 b^{2} + 2 b + 1 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 3 \cdot 1}{2 \cdot 3}

Simplify

2^{2} - 4 \cdot 3 \cdot 1 : 22 i

b_{1, 2} = \frac{- 2 \pm 2 2 i}{2 \cdot 3}

Separate the solutions

b_{1} = \frac{- 2 + 2 2 i}{2 \cdot 3}, b_{2} = \frac{- 2 - 2 2 i}{2 \cdot 3}

b = \frac{- 2 + 2 2 i}{2 \cdot 3} : - \frac{1}{3} + i \frac{2}{3}

b = \frac{- 2 - 2 2 i}{2 \cdot 3} : - \frac{1}{3} - i \frac{2}{3}

The solutions to the quadratic equation are:

b = - \frac{1}{3} + i \frac{2}{3}, b = - \frac{1}{3} - i \frac{2}{3}

x^{2} + 3 x - 68 = 3 x - 43 x - 4

9 w^{2} + 4 = 0

9 x - 4 x^{2} = 7

x^{2} - 18 x + 3 = 0

2 (x - 2)^{2} = 288

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