What are the solutions to the equation 3x^2+41x+n=0 ?

The solutions to the equation 3x^2+41x+n=0 are x=(-41+sqrt(1681-12n))/6 ,x=(-41-sqrt(1681-12n))/6

Find the zeros of 3x^2+41x+n=0

The zeros of 3x^2+41x+n=0 are x=(-41+sqrt(1681-12n))/6 ,x=(-41-sqrt(1681-12n))/6

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3 x^{2} + 41 x + n = 0

x = \frac{- 41 + 1681 - 12 n}{6}, x = \frac{- 41 - 1681 - 12 n}{6}

Solution steps

3 x^{2} + 41 x + n = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 41 \pm 4 1 ^{2} - 4 \cdot 3 n}{2 \cdot 3}

Simplify

4 1^{2} - 4 \cdot 3 n : 1681 - 12 n

x_{1, 2} = \frac{- 41 \pm 1681 - 12 n}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 41 + 1681 - 12 n}{2 \cdot 3}, x_{2} = \frac{- 41 - 1681 - 12 n}{2 \cdot 3}

x = \frac{- 41 + 1681 - 12 n}{2 \cdot 3} : \frac{- 41 + 1681 - 12 n}{6}

x = \frac{- 41 - 1681 - 12 n}{2 \cdot 3} : \frac{- 41 - 1681 - 12 n}{6}

The solutions to the quadratic equation are:

x = \frac{- 41 + 1681 - 12 n}{6}, x = \frac{- 41 - 1681 - 12 n}{6}

x^{2} + x - 50 = 0

56 b^{2} + 112 b - 102 = 3 + 7 b^{2}

x, \frac{x ^{2}}{25} + \frac{y ^{2}}{16} = 0

8 x^{2} + 8 x + 8 = 0

x^{2} - 4 = - 2 x - 1

What are the solutions to the equation 3x^2+41x+n=0 ?
The solutions to the equation 3x^2+41x+n=0 are x=(-41+sqrt(1681-12n))/6 ,x=(-41-sqrt(1681-12n))/6
Find the zeros of 3x^2+41x+n=0
The zeros of 3x^2+41x+n=0 are x=(-41+sqrt(1681-12n))/6 ,x=(-41-sqrt(1681-12n))/6