What are the solutions to the equation 4x^2+20x=0 ?

The solutions to the equation 4x^2+20x=0 are x=0,x=-5

Find the zeros of 4x^2+20x=0

The zeros of 4x^2+20x=0 are x=0,x=-5

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4 x^{2} + 20 x = 0

x = 0, x = - 5

Solution steps

4 x^{2} + 20 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 4 \cdot 0}{2 \cdot 4}

2 0^{2} - 4 \cdot 4 \cdot 0 = 20

x_{1, 2} = \frac{- 20 \pm 20}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 20 + 20}{2 \cdot 4}, x_{2} = \frac{- 20 - 20}{2 \cdot 4}

x = \frac{- 20 + 20}{2 \cdot 4} : 0

x = \frac{- 20 - 20}{2 \cdot 4} : - 5

The solutions to the quadratic equation are:

x = 0, x = - 5

\frac{10 !}{5}

4500

\frac{2 π}{\frac{9}{8}}

3 (5 x + 7)^{2} - 24 = - 12

4 a^{2} - 20 a + 25 = 0

What are the solutions to the equation 4x^2+20x=0 ?
The solutions to the equation 4x^2+20x=0 are x=0,x=-5
Find the zeros of 4x^2+20x=0
The zeros of 4x^2+20x=0 are x=0,x=-5