What are the solutions to the equation 4y=y^2-12 ?

The solutions to the equation 4y=y^2-12 are y=6,y=-2

Find the zeros of 4y=y^2-12

The zeros of 4y=y^2-12 are y=6,y=-2

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4 y = y^{2} - 12

y = 6, y = - 2

Solution steps

4 y = y^{2} - 12

Switch sides

y^{2} - 12 = 4 y

Move

4 y

to the left side

y^{2} - 12 - 4 y = 0

Write in the standard form

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

y^{2} - 4 y - 12 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 1 \cdot (- 12) = 8

y_{1, 2} = \frac{- ( - 4 ) \pm 8}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 4 ) + 8}{2 \cdot 1}, y_{2} = \frac{- ( - 4 ) - 8}{2 \cdot 1}

y = \frac{- ( - 4 ) + 8}{2 \cdot 1} : 6

y = \frac{- ( - 4 ) - 8}{2 \cdot 1} : - 2

The solutions to the quadratic equation are:

y = 6, y = - 2

(a + b)^{2} - (a - b)^{2}

3 p - 2 c^{3} - 3 c^{3} p + 2

\frac{4 x ^{2}}{25} - \frac{1}{25} = 0

6 x^{2} + 5 x = 4

3 x - 5 > 10

What are the solutions to the equation 4y=y^2-12 ?
The solutions to the equation 4y=y^2-12 are y=6,y=-2
Find the zeros of 4y=y^2-12
The zeros of 4y=y^2-12 are y=6,y=-2