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

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

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

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

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3 x^{2} + x = 400

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

Decimal

x = 11.38154 \dots, x = - 11.71487 \dots

Solution steps

3 x^{2} + x = 400

Move

400

to the left side

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

10 x + 18 + 2 x^{2} + 2 = 4 x^{2} + 28

y^{2} - 5 = 0

64 = 80 x - 16 x^{2}

(x + 2) (x - 6) + 101 = (2 x - 5) (x + 3)

(x - 3)^{2} + 8 = 17

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