What are the solutions to the equation 3b^2+4b+8=0 ?

The solutions to the equation 3b^2+4b+8=0 are b=-2/3+i(2sqrt(5))/3 ,b=-2/3-i(2sqrt(5))/3

Find the zeros of 3b^2+4b+8=0

The zeros of 3b^2+4b+8=0 are b=-2/3+i(2sqrt(5))/3 ,b=-2/3-i(2sqrt(5))/3

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3 b^{2} + 4 b + 8 = 0

b = - \frac{2}{3} + i \frac{2 5}{3}, b = - \frac{2}{3} - i \frac{2 5}{3}

Solution steps

3 b^{2} + 4 b + 8 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 3 \cdot 8}{2 \cdot 3}

Simplify

4^{2} - 4 \cdot 3 \cdot 8 : 45 i

b_{1, 2} = \frac{- 4 \pm 4 5 i}{2 \cdot 3}

Separate the solutions

b_{1} = \frac{- 4 + 4 5 i}{2 \cdot 3}, b_{2} = \frac{- 4 - 4 5 i}{2 \cdot 3}

b = \frac{- 4 + 4 5 i}{2 \cdot 3} : - \frac{2}{3} + i \frac{2 5}{3}

b = \frac{- 4 - 4 5 i}{2 \cdot 3} : - \frac{2}{3} - i \frac{2 5}{3}

The solutions to the quadratic equation are:

b = - \frac{2}{3} + i \frac{2 5}{3}, b = - \frac{2}{3} - i \frac{2 5}{3}

∣ x - 2 ∣ \geq 1

v + 2 \geq 6

(x - 4)^{2} - (x - 4) - 20 = 0

x^{2} - 9 x = - 20

4 x^{2} - 12 x y + 9 y^{2}

What are the solutions to the equation 3b^2+4b+8=0 ?
The solutions to the equation 3b^2+4b+8=0 are b=-2/3+i(2sqrt(5))/3 ,b=-2/3-i(2sqrt(5))/3
Find the zeros of 3b^2+4b+8=0
The zeros of 3b^2+4b+8=0 are b=-2/3+i(2sqrt(5))/3 ,b=-2/3-i(2sqrt(5))/3