What are the solutions to the equation b^2+2b=-22 ?

The solutions to the equation b^2+2b=-22 are b=-1+sqrt(21)i,b=-1-sqrt(21)i

Find the zeros of b^2+2b=-22

The zeros of b^2+2b=-22 are b=-1+sqrt(21)i,b=-1-sqrt(21)i

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b^{2} + 2 b = - 22

b = - 1 + 21 i, b = - 1 - 21 i

Solution steps

b^{2} + 2 b = - 22

Move

22

to the left side

b^{2} + 2 b + 22 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot 22}{2 \cdot 1}

Simplify

2^{2} - 4 \cdot 1 \cdot 22 : 221 i

b_{1, 2} = \frac{- 2 \pm 2 21 i}{2 \cdot 1}

Separate the solutions

b_{1} = \frac{- 2 + 2 21 i}{2 \cdot 1}, b_{2} = \frac{- 2 - 2 21 i}{2 \cdot 1}

b = \frac{- 2 + 2 21 i}{2 \cdot 1} : - 1 + 21 i

b = \frac{- 2 - 2 21 i}{2 \cdot 1} : - 1 - 21 i

The solutions to the quadratic equation are:

b = - 1 + 21 i, b = - 1 - 21 i

(3 x - 2)^{2} - 11 = 25

5 - 2 x^{2} = 4 x

x^{2} + 8 = - 41

28 = x^{2} - 4 x + 3 x - 2

- 6 - 12 n^{2} = - 396

What are the solutions to the equation b^2+2b=-22 ?
The solutions to the equation b^2+2b=-22 are b=-1+sqrt(21)i,b=-1-sqrt(21)i
Find the zeros of b^2+2b=-22
The zeros of b^2+2b=-22 are b=-1+sqrt(21)i,b=-1-sqrt(21)i