What are the solutions to the equation 12v^2+31v+20=0 ?

The solutions to the equation 12v^2+31v+20=0 are v=-5/4 ,v=-4/3

Find the zeros of 12v^2+31v+20=0

The zeros of 12v^2+31v+20=0 are v=-5/4 ,v=-4/3

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12 v^{2} + 31 v + 20 = 0

v = - \frac{5}{4}, v = - \frac{4}{3}

Decimal

v = - 1.25, v = - 1.33333 \dots

Solution steps

12 v^{2} + 31 v + 20 = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- 31 \pm 3 1 ^{2} - 4 \cdot 12 \cdot 20}{2 \cdot 12}

3 1^{2} - 4 \cdot 12 \cdot 20 = 1

v_{1, 2} = \frac{- 31 \pm 1}{2 \cdot 12}

Separate the solutions

v_{1} = \frac{- 31 + 1}{2 \cdot 12}, v_{2} = \frac{- 31 - 1}{2 \cdot 12}

v = \frac{- 31 + 1}{2 \cdot 12} : - \frac{5}{4}

v = \frac{- 31 - 1}{2 \cdot 12} : - \frac{4}{3}

The solutions to the quadratic equation are:

v = - \frac{5}{4}, v = - \frac{4}{3}

x, x^{2} + 2 x y + y^{2} = 1

x^{2} - 4 x + 5

3 x^{2} = 1

x^{2} - 30 x + 225 = 441

s^{2} - 2 s + 5

What are the solutions to the equation 12v^2+31v+20=0 ?
The solutions to the equation 12v^2+31v+20=0 are v=-5/4 ,v=-4/3
Find the zeros of 12v^2+31v+20=0
The zeros of 12v^2+31v+20=0 are v=-5/4 ,v=-4/3