What is 2<x^2+2x+2<= 10 ?

The solution to 2<x^2+2x+2<= 10 is -4<= x<-2\lor 0<x<= 2

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2<x^2+2x+2<= 10

2 < x^{2} + 2 x + 2 \leq 10

- 4 \leq x < - 2 or 0 < x \leq 2

Solution steps

2 < x^{2} + 2 x + 2 \leq 10

a < u \leq b

then

a < u and u \leq b

2 < x^{2} + 2 x + 2 and x^{2} + 2 x + 2 \leq 10

2 < x^{2} + 2 x + 2 : x < - 2 or x > 0

x^{2} + 2 x + 2 \leq 10 : - 4 \leq x \leq 2

Combine the intervals

(x < - 2 or x > 0) and - 4 \leq x \leq 2

Merge Overlapping Intervals

- 4 \leq x < - 2 or 0 < x \leq 2

(4 x - 4)^{2}

5 x^{2} + 14 x - 20 = 0

x^{2} + 2 x - 24 > 0

\frac{y ^{- 2} - x ^{- 2}}{y ^{- 2} + x ^{- 2}}

∣ x - 6 ∣ = 2

What is 2<x^2+2x+2<= 10 ?
The solution to 2<x^2+2x+2<= 10 is -4<= x<-2\lor 0<x<= 2