What are the solutions to the equation 12x^2+192x=0 ?

The solutions to the equation 12x^2+192x=0 are x=0,x=-16

Find the zeros of 12x^2+192x=0

The zeros of 12x^2+192x=0 are x=0,x=-16

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12 x^{2} + 192 x = 0

x = 0, x = - 16

Solution steps

12 x^{2} + 192 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 192 \pm 19 2 ^{2} - 4 \cdot 12 \cdot 0}{2 \cdot 12}

19 2^{2} - 4 \cdot 12 \cdot 0 = 192

x_{1, 2} = \frac{- 192 \pm 192}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- 192 + 192}{2 \cdot 12}, x_{2} = \frac{- 192 - 192}{2 \cdot 12}

x = \frac{- 192 + 192}{2 \cdot 12} : 0

x = \frac{- 192 - 192}{2 \cdot 12} : - 16

The solutions to the quadratic equation are:

x = 0, x = - 16

(x + 6)^{2} = - 16

y, x^{2} - y + 1 = x - y + 1

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

2 x^{2} - 18 x + 40 = 0

5 n^{2} - 30 = - 10

What are the solutions to the equation 12x^2+192x=0 ?
The solutions to the equation 12x^2+192x=0 are x=0,x=-16
Find the zeros of 12x^2+192x=0
The zeros of 12x^2+192x=0 are x=0,x=-16