What are the solutions to the equation b^2+2b=-20 ?

The solutions to the equation b^2+2b=-20 are b=-1+sqrt(19)i,b=-1-sqrt(19)i

Find the zeros of b^2+2b=-20

The zeros of b^2+2b=-20 are b=-1+sqrt(19)i,b=-1-sqrt(19)i

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b^{2} + 2 b = - 20

b = - 1 + 19 i, b = - 1 - 19 i

Solution steps

b^{2} + 2 b = - 20

Move

20

to the left side

b^{2} + 2 b + 20 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot 20}{2 \cdot 1}

Simplify

2^{2} - 4 \cdot 1 \cdot 20 : 219 i

b_{1, 2} = \frac{- 2 \pm 2 19 i}{2 \cdot 1}

Separate the solutions

b_{1} = \frac{- 2 + 2 19 i}{2 \cdot 1}, b_{2} = \frac{- 2 - 2 19 i}{2 \cdot 1}

b = \frac{- 2 + 2 19 i}{2 \cdot 1} : - 1 + 19 i

b = \frac{- 2 - 2 19 i}{2 \cdot 1} : - 1 - 19 i

The solutions to the quadratic equation are:

b = - 1 + 19 i, b = - 1 - 19 i

0.24 x - 0.38 (x + 2) = - 0.34 (x + 4)

\frac{1}{2} 29500 \cdot 0.82 5^{2} = \frac{1}{2} (29500 + 115000) x^{2}

(7 x + 1)^{2} = 5

\frac{x}{2} + \frac{5 x}{4} = \frac{x}{12}

- 1 + \frac{2}{3}

What are the solutions to the equation b^2+2b=-20 ?
The solutions to the equation b^2+2b=-20 are b=-1+sqrt(19)i,b=-1-sqrt(19)i
Find the zeros of b^2+2b=-20
The zeros of b^2+2b=-20 are b=-1+sqrt(19)i,b=-1-sqrt(19)i