What are the solutions to the equation solvefor x,12x^2+px=0 ?

The solutions to the equation solvefor x,12x^2+px=0 are x=0,x=-p/(12)

Find the zeros of solvefor x,12x^2+px=0

The zeros of solvefor x,12x^2+px=0 are x=0,x=-p/(12)

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solve for

x, 12 x^{2} + p x = 0

x = 0, x = - \frac{p}{12}

Solution steps

12 x^{2} + p x = 0

Solve with the quadratic formula

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

Simplify

p^{2} - 4 \cdot 12 \cdot 0 : p

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

Separate the solutions

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

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

x = \frac{- p - p}{2 \cdot 12} : - \frac{p}{12}

The solutions to the quadratic equation are:

x = 0, x = - \frac{p}{12}

4 x^{2} + 15 x = 0

(x + 1) (x^{2} - x + 1) = (x - 1)^{3}

x, g (x + 2) = x^{2} - 10 x - 3

x (2 x - 9) = 0

3 y^{2} + 7 y + 10 = 3

What are the solutions to the equation solvefor x,12x^2+px=0 ?
The solutions to the equation solvefor x,12x^2+px=0 are x=0,x=-p/(12)
Find the zeros of solvefor x,12x^2+px=0
The zeros of solvefor x,12x^2+px=0 are x=0,x=-p/(12)