What is 0<=-2x^2+8x+8<= 10 ?

The solution to 0<=-2x^2+8x+8<= 10 is -2sqrt(2)+2<= x<=-sqrt(3)+2\lor sqrt(3)+2<= x<= 2sqrt(2)+2

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0<=-2x^2+8x+8<= 10

0 \leq - 2 x^{2} + 8 x + 8 \leq 10

- 22 + 2 \leq x \leq - 3 + 2 or 3 + 2 \leq x \leq 22 + 2

Solution steps

0 \leq - 2 x^{2} + 8 x + 8 \leq 10

a \leq u \leq b

then

a \leq u and u \leq b

0 \leq - 2 x^{2} + 8 x + 8 and - 2 x^{2} + 8 x + 8 \leq 10

0 \leq - 2 x^{2} + 8 x + 8 : - 22 + 2 \leq x \leq 22 + 2

- 2 x^{2} + 8 x + 8 \leq 10 : x \leq - 3 + 2 or x \geq 3 + 2

Combine the intervals

- 22 + 2 \leq x \leq 22 + 2 and (x \leq - 3 + 2 or x \geq 3 + 2)

Merge Overlapping Intervals

- 22 + 2 \leq x \leq - 3 + 2 or 3 + 2 \leq x \leq 22 + 2

(3 - 2 i) (4 + i)

(x - 4) (x + 2) = 16

(x + 1)^{2} (x - 3) (x - 8) < 0

9 x^{2} - 7 = 0

36 x^{2} - 49 y^{2}

What is 0<=-2x^2+8x+8<= 10 ?
The solution to 0<=-2x^2+8x+8<= 10 is -2sqrt(2)+2<= x<=-sqrt(3)+2\lor sqrt(3)+2<= x<= 2sqrt(2)+2