What are the solutions to the equation 3u^2+24u=0 ?

The solutions to the equation 3u^2+24u=0 are u=0,u=-8

Find the zeros of 3u^2+24u=0

The zeros of 3u^2+24u=0 are u=0,u=-8

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3 u^{2} + 24 u = 0

u = 0, u = - 8

Solution steps

3 u^{2} + 24 u = 0

Solve with the quadratic formula

u_{1, 2} = \frac{- 24 \pm 2 4 ^{2} - 4 \cdot 3 \cdot 0}{2 \cdot 3}

2 4^{2} - 4 \cdot 3 \cdot 0 = 24

u_{1, 2} = \frac{- 24 \pm 24}{2 \cdot 3}

Separate the solutions

u_{1} = \frac{- 24 + 24}{2 \cdot 3}, u_{2} = \frac{- 24 - 24}{2 \cdot 3}

u = \frac{- 24 + 24}{2 \cdot 3} : 0

u = \frac{- 24 - 24}{2 \cdot 3} : - 8

The solutions to the quadratic equation are:

u = 0, u = - 8

16 x^{2} = - 56 x - 18

x^{2} + 5 x + 1 = 0

t, - 4.905 t^{2} - 0.517 t - 15 = 0

k^{2} + 5 k = 14

x^{2} - 4 x - 70 = 80 + x

What are the solutions to the equation 3u^2+24u=0 ?
The solutions to the equation 3u^2+24u=0 are u=0,u=-8
Find the zeros of 3u^2+24u=0
The zeros of 3u^2+24u=0 are u=0,u=-8