What are the solutions to the equation 4u^2+28u=0 ?

The solutions to the equation 4u^2+28u=0 are u=0,u=-7

Find the zeros of 4u^2+28u=0

The zeros of 4u^2+28u=0 are u=0,u=-7

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4 u^{2} + 28 u = 0

u = 0, u = - 7

Solution steps

4 u^{2} + 28 u = 0

Solve with the quadratic formula

u_{1, 2} = \frac{- 28 \pm 2 8 ^{2} - 4 \cdot 4 \cdot 0}{2 \cdot 4}

2 8^{2} - 4 \cdot 4 \cdot 0 = 28

u_{1, 2} = \frac{- 28 \pm 28}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- 28 + 28}{2 \cdot 4}, u_{2} = \frac{- 28 - 28}{2 \cdot 4}

u = \frac{- 28 + 28}{2 \cdot 4} : 0

u = \frac{- 28 - 28}{2 \cdot 4} : - 7

The solutions to the quadratic equation are:

u = 0, u = - 7

- 4 ∣ 3 x + 4 ∣ - 3 = - 15

(2 x + 5) (2 x - 3) \geq 3 x - 12

35 x^{2} + 25

10 = 8 h

- 95 = - 10 x + 5

What are the solutions to the equation 4u^2+28u=0 ?
The solutions to the equation 4u^2+28u=0 are u=0,u=-7
Find the zeros of 4u^2+28u=0
The zeros of 4u^2+28u=0 are u=0,u=-7