What are the solutions to the equation 3u^2+18u=0 ?

The solutions to the equation 3u^2+18u=0 are u=0,u=-6

Find the zeros of 3u^2+18u=0

The zeros of 3u^2+18u=0 are u=0,u=-6

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3 u^{2} + 18 u = 0

u = 0, u = - 6

Solution steps

3 u^{2} + 18 u = 0

Solve with the quadratic formula

u_{1, 2} = \frac{- 18 \pm 1 8 ^{2} - 4 \cdot 3 \cdot 0}{2 \cdot 3}

1 8^{2} - 4 \cdot 3 \cdot 0 = 18

u_{1, 2} = \frac{- 18 \pm 18}{2 \cdot 3}

Separate the solutions

u_{1} = \frac{- 18 + 18}{2 \cdot 3}, u_{2} = \frac{- 18 - 18}{2 \cdot 3}

u = \frac{- 18 + 18}{2 \cdot 3} : 0

u = \frac{- 18 - 18}{2 \cdot 3} : - 6

The solutions to the quadratic equation are:

u = 0, u = - 6

0.585 = - 0.26 x^{2} + 0.93 x + 0.8791

24 = 3 x^{2} + 34 x

x^{2} - 8 x + 10 = 3

4 x^{2} - 3 = - 11 x

2 x - 1 = - x^{2}

What are the solutions to the equation 3u^2+18u=0 ?
The solutions to the equation 3u^2+18u=0 are u=0,u=-6
Find the zeros of 3u^2+18u=0
The zeros of 3u^2+18u=0 are u=0,u=-6