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

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

Solve for x

$$x\in \left(-5,0\right)$$

Solution Steps

Use the distributive property to multiply -2 by x^{2}+3x.

$$x^{2}+x-2x^{2}-6x>0$$

Combine x^{2} and -2x^{2} to get -x^{2}.

$$-x^{2}+x-6x>0$$

Combine x and -6x to get -5x.

$$-x^{2}-5x>0$$

Multiply the inequality by -1 to make the coefficient of the highest power in -x^{2}-5x positive. Since -1 is negative, the inequality direction is changed.

$$x^{2}+5x<0$$

Factor out x.

$$x\left(x+5\right)<0$$

For the product to be negative, x+5 and x have to be of the opposite signs. Consider the case when x+5 is positive and x is negative.

$$x+5>0 x<0$$

The solution satisfying both inequalities is x\in \left(-5,0\right).

$$x\in \left(-5,0\right)$$

Consider the case when x is positive and x+5 is negative.

$$x>0 x+5<0$$

This is false for any x.

$$x\in \emptyset$$

The final solution is the union of the obtained solutions.

$$x\in \left(-5,0\right)$$