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

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

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

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

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

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

Decimal Notation

k = 1.59066 \dots, k = - 1.25733 \dots

Solution steps

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

Move

6

to the left side

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

x^{2} - 5 x + 3 = 0

(3 k + 4)^{2} + 15 = 0

y^{2} + 8 = 33

4 x^{2} - 5 = x^{2}

c x^{2} + 12 x + c = 0

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