What are the solutions to the equation 3n^2+n=400 ?

The solutions to the equation 3n^2+n=400 are n=(-1+sqrt(4801))/6 ,n=(-1-sqrt(4801))/6

Find the zeros of 3n^2+n=400

The zeros of 3n^2+n=400 are n=(-1+sqrt(4801))/6 ,n=(-1-sqrt(4801))/6

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3 n^{2} + n = 400

n = \frac{- 1 + 4801}{6}, n = \frac{- 1 - 4801}{6}

Decimal Notation

n = 11.38154 \dots, n = - 11.71487 \dots

Solution steps

3 n^{2} + n = 400

Move

400

to the left side

3 n^{2} + n - 400 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 3 ( - 400 )}{2 \cdot 3}

1^{2} - 4 \cdot 3 (- 400) = 4801

n_{1, 2} = \frac{- 1 \pm 4801}{2 \cdot 3}

Separate the solutions

n_{1} = \frac{- 1 + 4801}{2 \cdot 3}, n_{2} = \frac{- 1 - 4801}{2 \cdot 3}

n = \frac{- 1 + 4801}{2 \cdot 3} : \frac{- 1 + 4801}{6}

n = \frac{- 1 - 4801}{2 \cdot 3} : \frac{- 1 - 4801}{6}

The solutions to the quadratic equation are:

n = \frac{- 1 + 4801}{6}, n = \frac{- 1 - 4801}{6}

1 = x^{2} - 6

3 r^{2} + r - 9 = - 2

5 (x^{2} + 2) - 8 = 7

x^{2} + 10 x - 4

3 y^{2} - y + 4 = 0

What are the solutions to the equation 3n^2+n=400 ?
The solutions to the equation 3n^2+n=400 are n=(-1+sqrt(4801))/6 ,n=(-1-sqrt(4801))/6
Find the zeros of 3n^2+n=400
The zeros of 3n^2+n=400 are n=(-1+sqrt(4801))/6 ,n=(-1-sqrt(4801))/6