What are the solutions to the equation 3n^2-n-6=0 ?

The solutions to the equation 3n^2-n-6=0 are n=(1+sqrt(73))/6 ,n=(1-sqrt(73))/6

Find the zeros of 3n^2-n-6=0

The zeros of 3n^2-n-6=0 are n=(1+sqrt(73))/6 ,n=(1-sqrt(73))/6

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

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

Decimal Notation

n = 1.59066 \dots, n = - 1.25733 \dots

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

s^{2} - s = 0

16.7 - \frac{20 x ^{2}}{10} = 0

- x^{2} + 10 x + 25 = 0

10 = 3 Q^{2} - 18 Q + 5

s^{2} + 4 s + 13

What are the solutions to the equation 3n^2-n-6=0 ?
The solutions to the equation 3n^2-n-6=0 are n=(1+sqrt(73))/6 ,n=(1-sqrt(73))/6
Find the zeros of 3n^2-n-6=0
The zeros of 3n^2-n-6=0 are n=(1+sqrt(73))/6 ,n=(1-sqrt(73))/6