What are the solutions to the equation 3n^2+11n+4=0 ?

The solutions to the equation 3n^2+11n+4=0 are n=(-11+sqrt(73))/6 ,n=(-11-sqrt(73))/6

Find the zeros of 3n^2+11n+4=0

The zeros of 3n^2+11n+4=0 are n=(-11+sqrt(73))/6 ,n=(-11-sqrt(73))/6

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3 n^{2} + 11 n + 4 = 0

n = \frac{- 11 + 73}{6}, n = \frac{- 11 - 73}{6}

Solution steps

3 n^{2} + 11 n + 4 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 3 \cdot 4}{2 \cdot 3}

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

n_{1, 2} = \frac{- 11 \pm 73}{2 \cdot 3}

Separate the solutions

n_{1} = \frac{- 11 + 73}{2 \cdot 3}, n_{2} = \frac{- 11 - 73}{2 \cdot 3}

n = \frac{- 11 + 73}{2 \cdot 3} : \frac{- 11 + 73}{6}

n = \frac{- 11 - 73}{2 \cdot 3} : \frac{- 11 - 73}{6}

The solutions to the quadratic equation are:

n = \frac{- 11 + 73}{6}, n = \frac{- 11 - 73}{6}

8 x^{2} = 10 x

\frac{60}{10}

x^{2} - 2 x - 9 > 6

x - 15 = 12

\frac{3}{2} t^{2} = 5 t + \frac{5}{7} t^{2}

What are the solutions to the equation 3n^2+11n+4=0 ?
The solutions to the equation 3n^2+11n+4=0 are n=(-11+sqrt(73))/6 ,n=(-11-sqrt(73))/6
Find the zeros of 3n^2+11n+4=0
The zeros of 3n^2+11n+4=0 are n=(-11+sqrt(73))/6 ,n=(-11-sqrt(73))/6