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

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

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

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

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

v = - \frac{3}{2}

Decimal Notation

v = - 1.5

Solution steps

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

Move

12 v

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

v_{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

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

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

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

v = - \frac{3}{2}

The solution to the quadratic equation is:

v = - \frac{3}{2}

\frac{1}{1 - \frac{e}{10}}

\frac{10}{X ^{2} - 16} \geq 0

36 \cdot 2

10 x^{2} + 9 x + 2

\frac{d}{d y} ((y - 1)^{2} (2 - y) (y - 3))

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