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

The solutions to the equation b^2=-3b are b=0,b=-3

Find the zeros of b^2=-3b

The zeros of b^2=-3b are b=0,b=-3

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b^{2} = - 3 b

b = 0, b = - 3

Solution steps

b^{2} = - 3 b

Move

3 b

to the left side

b^{2} + 3 b = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot 0 = 3

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

Separate the solutions

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

b = \frac{- 3 + 3}{2 \cdot 1} : 0

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

The solutions to the quadratic equation are:

b = 0, b = - 3

- 3 (y - 2) \geq - 9

\frac{4.2 x ^{4} y ^{14}}{0.6 x ^{9} y ^{5}}

- 5 x < - 15

(6 + 5 i) (6 - 5 i)

c^{2} + 11 c = 12

What are the solutions to the equation b^2=-3b ?
The solutions to the equation b^2=-3b are b=0,b=-3
Find the zeros of b^2=-3b
The zeros of b^2=-3b are b=0,b=-3