What are the solutions to the equation 3v^2-v-12=-6 ?

The solutions to the equation 3v^2-v-12=-6 are v=(1+sqrt(73))/6 ,v=(1-sqrt(73))/6

Find the zeros of 3v^2-v-12=-6

The zeros of 3v^2-v-12=-6 are v=(1+sqrt(73))/6 ,v=(1-sqrt(73))/6

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3 v^{2} - v - 12 = - 6

v = \frac{1 + 73}{6}, v = \frac{1 - 73}{6}

Decimal Notation

v = 1.59066 \dots, v = - 1.25733 \dots

Solution steps

3 v^{2} - v - 12 = - 6

Move

6

to the left side

3 v^{2} - v - 6 = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 3 ( - 6 )}{2 \cdot 3}

(- 1)^{2} - 4 \cdot 3 (- 6) = 73

v_{1, 2} = \frac{- ( - 1 ) \pm 73}{2 \cdot 3}

Separate the solutions

v_{1} = \frac{- ( - 1 ) + 73}{2 \cdot 3}, v_{2} = \frac{- ( - 1 ) - 73}{2 \cdot 3}

v = \frac{- ( - 1 ) + 73}{2 \cdot 3} : \frac{1 + 73}{6}

v = \frac{- ( - 1 ) - 73}{2 \cdot 3} : \frac{1 - 73}{6}

The solutions to the quadratic equation are:

v = \frac{1 + 73}{6}, v = \frac{1 - 73}{6}

x^{2} - 16 x = - 73

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

30 x^{2} - 5 x = 0

x^{2} + 4 x + 40

96 = 80 x - 16 x^{2}

What are the solutions to the equation 3v^2-v-12=-6 ?
The solutions to the equation 3v^2-v-12=-6 are v=(1+sqrt(73))/6 ,v=(1-sqrt(73))/6
Find the zeros of 3v^2-v-12=-6
The zeros of 3v^2-v-12=-6 are v=(1+sqrt(73))/6 ,v=(1-sqrt(73))/6