What are the solutions to the equation 3u^2+27u=0 ?

The solutions to the equation 3u^2+27u=0 are u=0,u=-9

Find the zeros of 3u^2+27u=0

The zeros of 3u^2+27u=0 are u=0,u=-9

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3 u^{2} + 27 u = 0

u = 0, u = - 9

Solution steps

3 u^{2} + 27 u = 0

Solve with the quadratic formula

u_{1, 2} = \frac{- 27 \pm 2 7 ^{2} - 4 \cdot 3 \cdot 0}{2 \cdot 3}

2 7^{2} - 4 \cdot 3 \cdot 0 = 27

u_{1, 2} = \frac{- 27 \pm 27}{2 \cdot 3}

Separate the solutions

u_{1} = \frac{- 27 + 27}{2 \cdot 3}, u_{2} = \frac{- 27 - 27}{2 \cdot 3}

u = \frac{- 27 + 27}{2 \cdot 3} : 0

u = \frac{- 27 - 27}{2 \cdot 3} : - 9

The solutions to the quadratic equation are:

u = 0, u = - 9

x^{2} + 6 x + 9 = 0

x, 9 x^{2} + 4 y^{2} + 36 x - 24 y + 72 = 0

x, 4 x^{2} - y^{2} + 2 z^{2} + 4 = 0

8 x^{2} + 12 x + 3 = 0

x^{2} + 6 x = 13

What are the solutions to the equation 3u^2+27u=0 ?
The solutions to the equation 3u^2+27u=0 are u=0,u=-9
Find the zeros of 3u^2+27u=0
The zeros of 3u^2+27u=0 are u=0,u=-9