What are the solutions to the equation v^2=-19v ?

The solutions to the equation v^2=-19v are v=0,v=-19

Find the zeros of v^2=-19v

The zeros of v^2=-19v are v=0,v=-19

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v^{2} = - 19 v

v = 0, v = - 19

Solution steps

v^{2} = - 19 v

Move

19 v

to the left side

v^{2} + 19 v = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- 19 \pm 1 9 ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

1 9^{2} - 4 \cdot 1 \cdot 0 = 19

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

Separate the solutions

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

v = \frac{- 19 + 19}{2 \cdot 1} : 0

v = \frac{- 19 - 19}{2 \cdot 1} : - 19

The solutions to the quadratic equation are:

v = 0, v = - 19

0 = x^{2} + 16 x + 64

N (d) = - 2 d^{2} + 16 d - 14

0.8 x^{2} - 8.16 x + 4.8 = 0

8 x^{2} - 2 x - 21 = 0

a, a^{5} + b^{5} = (a + b) (a^{4} - a^{3} b + a^{2} b^{2} - a b^{3} + b^{4})

What are the solutions to the equation v^2=-19v ?
The solutions to the equation v^2=-19v are v=0,v=-19
Find the zeros of v^2=-19v
The zeros of v^2=-19v are v=0,v=-19