What is x^2+x-2(x^2+3x)>0 ?

The solution to x^2+x-2(x^2+3x)>0 is -5<x<0

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x^2+x-2(x^2+3x)>0

x^{2} + x - 2 (x^{2} + 3 x) > 0

- 5 < x < 0

Solution steps

x^{2} + x - 2 (x^{2} + 3 x) > 0

Factor

x^{2} + x - 2 (x^{2} + 3 x) : - x (x + 5)

- x (x + 5) > 0

Multiply both sides by

- 1

(reverse the inequality)

(- x (x + 5)) (- 1) < 0 \cdot (- 1)

Simplify

x (x + 5) < 0

Identify the intervals

- 5 < x < 0

K (r, s) = 7 r - s

1 + 0 = c

12 x^{2} - 30 x - 18

y^{2} - 2 y + 10 = 0

2^{(\frac{1}{2})}