What are the solutions to the equation 4v^2=-12v-5 ?

The solutions to the equation 4v^2=-12v-5 are v=-1/2 ,v=-5/2

Find the zeros of 4v^2=-12v-5

The zeros of 4v^2=-12v-5 are v=-1/2 ,v=-5/2

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4 v^{2} = - 12 v - 5

v = - \frac{1}{2}, v = - \frac{5}{2}

Solution steps

4 v^{2} = - 12 v - 5

Move

5

to the left side

4 v^{2} + 5 = - 12 v

Move

12 v

to the left side

4 v^{2} + 5 + 12 v = 0

Write in the standard form

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

4 v^{2} + 12 v + 5 = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 4 \cdot 5}{2 \cdot 4}

1 2^{2} - 4 \cdot 4 \cdot 5 = 8

v_{1, 2} = \frac{- 12 \pm 8}{2 \cdot 4}

Separate the solutions

v_{1} = \frac{- 12 + 8}{2 \cdot 4}, v_{2} = \frac{- 12 - 8}{2 \cdot 4}

v = \frac{- 12 + 8}{2 \cdot 4} : - \frac{1}{2}

v = \frac{- 12 - 8}{2 \cdot 4} : - \frac{5}{2}

The solutions to the quadratic equation are:

v = - \frac{1}{2}, v = - \frac{5}{2}

x^{2} + 4 x - 12 = 0

15 x^{2} + 1.5 x + 1 = 0

- x^{2} - 2 x + 48 = 3 x + 12

9 x^{2} - 3 = 0

9 (x^{2} - 11) - 8 = 14

What are the solutions to the equation 4v^2=-12v-5 ?
The solutions to the equation 4v^2=-12v-5 are v=-1/2 ,v=-5/2
Find the zeros of 4v^2=-12v-5
The zeros of 4v^2=-12v-5 are v=-1/2 ,v=-5/2