What are the solutions to the equation x(-x+2}{12}=\frac{5x)/2 ?

The solutions to the equation x(-x+2}{12}=\frac{5x)/2 are x=-28,x=0

Find the zeros of x(-x+2}{12}=\frac{5x)/2

The zeros of x(-x+2}{12}=\frac{5x)/2 are x=-28,x=0

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x \frac{- x + 2}{12} = \frac{5 x}{2}

x = - 28, x = 0

Solution steps

x \frac{- x + 2}{12} = \frac{5 x}{2}

Multiply both sides by

2

\frac{x ( - x + 2 )}{6} = 5 x

Expand

\frac{x ( - x + 2 )}{6} : - \frac{x ^{2}}{6} + \frac{x}{3}

- \frac{x ^{2}}{6} + \frac{x}{3} = 5 x

Move

5 x

to the left side

\frac{- x ^{2} - 28 x}{6} = 0

Write in the standard form

a x^{2} + b x + c = 0

- \frac{x ^{2}}{6} - \frac{14 x}{3} = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - \frac{14}{3} ) \pm ( - \frac{14}{3} ) ^{2} - 4 ( - \frac{1}{6} ) \cdot 0}{2 ( - \frac{1}{6} )}

(- \frac{14}{3})^{2} - 4 (- \frac{1}{6}) \cdot 0 = \frac{14}{3}

x_{1, 2} = \frac{- ( - \frac{14}{3} ) \pm \frac{14}{3}}{2 ( - \frac{1}{6} )}

Separate the solutions

x_{1} = \frac{- ( - \frac{14}{3} ) + \frac{14}{3}}{2 ( - \frac{1}{6} )}, x_{2} = \frac{- ( - \frac{14}{3} ) - \frac{14}{3}}{2 ( - \frac{1}{6} )}

x = \frac{- ( - \frac{14}{3} ) + \frac{14}{3}}{2 ( - \frac{1}{6} )} : - 28

x = \frac{- ( - \frac{14}{3} ) - \frac{14}{3}}{2 ( - \frac{1}{6} )} : 0

The solutions to the quadratic equation are:

x = - 28, x = 0

x, f = in (x^{2} + 1)

x^{2} - x - 6 = 0

8 x^{2} = 800

3 x^{2} - 3 x + 3 = 0

- m^{2} - 9 = 0

What are the solutions to the equation x(-x+2}{12}=\frac{5x)/2 ?
The solutions to the equation x(-x+2}{12}=\frac{5x)/2 are x=-28,x=0
Find the zeros of x(-x+2}{12}=\frac{5x)/2
The zeros of x(-x+2}{12}=\frac{5x)/2 are x=-28,x=0