What are the solutions to the equation -12y=4y^2+9 ?

The solutions to the equation -12y=4y^2+9 are y=-3/2

Find the zeros of -12y=4y^2+9

The zeros of -12y=4y^2+9 are y=-3/2

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- 12 y = 4 y^{2} + 9

y = - \frac{3}{2}

Decimal Notation

y = - 1.5

Solution steps

- 12 y = 4 y^{2} + 9

Switch sides

4 y^{2} + 9 = - 12 y

Move

12 y

to the left side

4 y^{2} + 9 + 12 y = 0

Write in the standard form

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

4 y^{2} + 12 y + 9 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 4 \cdot 9}{2 \cdot 4}

1 2^{2} - 4 \cdot 4 \cdot 9 = 0

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

y = \frac{- 12}{2 \cdot 4}

\frac{- 12}{2 \cdot 4} = - \frac{3}{2}

y = - \frac{3}{2}

The solution to the quadratic equation is:

y = - \frac{3}{2}

x^{2} + 8 x + 16 = - 4

x^{2} - 50 x + 106 = 0

m^{2} - m - 1 = 0

(x + 3)^{2} = - 22

x^{2} - x - 380 = 0

What are the solutions to the equation -12y=4y^2+9 ?
The solutions to the equation -12y=4y^2+9 are y=-3/2
Find the zeros of -12y=4y^2+9
The zeros of -12y=4y^2+9 are y=-3/2