What are the solutions to the equation x^2+2000x=2001 ?

The solutions to the equation x^2+2000x=2001 are x=1,x=-2001

Find the zeros of x^2+2000x=2001

The zeros of x^2+2000x=2001 are x=1,x=-2001

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x^{2} + 2000 x = 2001

x = 1, x = - 2001

Solution steps

x^{2} + 2000 x = 2001

Move

2001

to the left side

x^{2} + 2000 x - 2001 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2000 \pm 200 0 ^{2} - 4 \cdot 1 \cdot ( - 2001 )}{2 \cdot 1}

200 0^{2} - 4 \cdot 1 \cdot (- 2001) = 2002

x_{1, 2} = \frac{- 2000 \pm 2002}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 2000 + 2002}{2 \cdot 1}, x_{2} = \frac{- 2000 - 2002}{2 \cdot 1}

x = \frac{- 2000 + 2002}{2 \cdot 1} : 1

x = \frac{- 2000 - 2002}{2 \cdot 1} : - 2001

The solutions to the quadratic equation are:

x = 1, x = - 2001

96 = 2 y^{2}

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

(9.81) (280) + \frac{1}{2} (5)^{2} = \frac{1}{2} x^{2}

(y - 1)^{2} = 1

u, 3 u^{2} - u - 3 = 0

What are the solutions to the equation x^2+2000x=2001 ?
The solutions to the equation x^2+2000x=2001 are x=1,x=-2001
Find the zeros of x^2+2000x=2001
The zeros of x^2+2000x=2001 are x=1,x=-2001