What are the solutions to the equation 1000=x^2+3x ?

The solutions to the equation 1000=x^2+3x are x=(-3+sqrt(4009))/2 ,x=(-3-sqrt(4009))/2

Find the zeros of 1000=x^2+3x

The zeros of 1000=x^2+3x are x=(-3+sqrt(4009))/2 ,x=(-3-sqrt(4009))/2

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1000 = x^{2} + 3 x

x = \frac{- 3 + 4009}{2}, x = \frac{- 3 - 4009}{2}

Solution steps

1000 = x^{2} + 3 x

Switch sides

x^{2} + 3 x = 1000

Move

1000

to the left side

x^{2} + 3 x - 1000 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot ( - 1000 )}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot (- 1000) = 4009

x_{1, 2} = \frac{- 3 \pm 4009}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 3 + 4009}{2 \cdot 1}, x_{2} = \frac{- 3 - 4009}{2 \cdot 1}

x = \frac{- 3 + 4009}{2 \cdot 1} : \frac{- 3 + 4009}{2}

x = \frac{- 3 - 4009}{2 \cdot 1} : \frac{- 3 - 4009}{2}

The solutions to the quadratic equation are:

x = \frac{- 3 + 4009}{2}, x = \frac{- 3 - 4009}{2}

\frac{2}{5} x = 14

2 x^{2} - 5 x + 2 \geq 0

\frac{7}{9} v + 16 = 27

\frac{1}{3} (x + 6) = \frac{2}{3} (x - 9)

4 m + \frac{3}{8} = \frac{67}{8}

What are the solutions to the equation 1000=x^2+3x ?
The solutions to the equation 1000=x^2+3x are x=(-3+sqrt(4009))/2 ,x=(-3-sqrt(4009))/2
Find the zeros of 1000=x^2+3x
The zeros of 1000=x^2+3x are x=(-3+sqrt(4009))/2 ,x=(-3-sqrt(4009))/2