What are the solutions to the equation y^2=12y ?

The solutions to the equation y^2=12y are y=12,y=0

Find the zeros of y^2=12y

The zeros of y^2=12y are y=12,y=0

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y^{2} = 12 y

y = 12, y = 0

Solution steps

y^{2} = 12 y

Move

12 y

to the left side

y^{2} - 12 y = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

(- 12)^{2} - 4 \cdot 1 \cdot 0 = 12

y_{1, 2} = \frac{- ( - 12 ) \pm 12}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 12 ) + 12}{2 \cdot 1}, y_{2} = \frac{- ( - 12 ) - 12}{2 \cdot 1}

y = \frac{- ( - 12 ) + 12}{2 \cdot 1} : 12

y = \frac{- ( - 12 ) - 12}{2 \cdot 1} : 0

The solutions to the quadratic equation are:

y = 12, y = 0

3 5^{2}

- \frac{2}{12} + (- \frac{4}{6}) + \frac{5}{3}

25 - 16 x^{2} \geq 0

\frac{1}{2} + \frac{12}{x ^{2}} > \frac{5}{x}

3 x + 4 = 16

What are the solutions to the equation y^2=12y ?
The solutions to the equation y^2=12y are y=12,y=0
Find the zeros of y^2=12y
The zeros of y^2=12y are y=12,y=0