What is x^2+2x-2>2000 ?

The solution to x^2+2x-2>2000 is x sqrt(2003)-1

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x^2+2x-2>2000

x^{2} + 2 x - 2 > 2000

x < - 2003 - 1 or x > 2003 - 1

Interval Notation

(- \infty, - 2003 - 1) \cup (2003 - 1, \infty)

Decimal Notation

x < - 45.75488 \dots or x > 43.75488 \dots

Solution steps

x^{2} + 2 x - 2 > 2000

Rewrite in standard form

x^{2} + 2 x - 2002 > 0

Complete the square

x^{2} + 2 x - 2002 : (x + 1)^{2} - 2003

(x + 1)^{2} - 2003 > 0

Move

2003

to the right side

(x + 1)^{2} > 2003

For

u^{n} > a

, if

n

is even then

x + 1 < - 2003 or x + 1 > 2003

x + 1 < - 2003 : x < - 2003 - 1

x + 1 > 2003 : x > 2003 - 1

Combine the intervals

x < - 2003 - 1 or x > 2003 - 1

2 u^{2} + 2 = - 5 u

x - 3 = - (x - 4)^{2} + 5

0.0004

\frac{\frac{3 !}{2 ! 1 !}}{3 6 ^{2}}

2 l + 2 w

What is x^2+2x-2>2000 ?
The solution to x^2+2x-2>2000 is x<-sqrt(2003)-1\lor x>sqrt(2003)-1