Solve -(x-3)(x+1)<0

$$- ( x - 3 ) ( x + 1 ) < 0$$

Solve for x

$$x\in \left(-\infty,-1\right)\cup \left(3,\infty\right)$$

Steps Using the Quadratic Formula

To find the opposite of x-3, find the opposite of each term.

$$\left(-x-\left(-3\right)\right)\left(x+1\right)<0$$

The opposite of -3 is 3.

$$\left(-x+3\right)\left(x+1\right)<0$$

Apply the distributive property by multiplying each term of -x+3 by each term of x+1.

$$-x^{2}-x+3x+3<0$$

Combine -x and 3x to get 2x.

$$-x^{2}+2x+3<0$$

Multiply the inequality by -1 to make the coefficient of the highest power in -x^{2}+2x+3 positive. Since -1 is negative, the inequality direction is changed.

$$x^{2}-2x-3>0$$

To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$x^{2}-2x-3=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, -2 for b, and -3 for c in the quadratic formula.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 1\left(-3\right)}}{2}$$

Do the calculations.

$$x=\frac{2±4}{2}$$

Solve the equation x=\frac{2±4}{2} when ± is plus and when ± is minus.

$$x=3 x=-1$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-3\right)\left(x+1\right)>0$$

For the product to be positive, x-3 and x+1 have to be both negative or both positive. Consider the case when x-3 and x+1 are both negative.

$$x-3<0 x+1<0$$

The solution satisfying both inequalities is x<-1.

$$x<-1$$

Consider the case when x-3 and x+1 are both positive.

$$x+1>0 x-3>0$$

The solution satisfying both inequalities is x>3.

$$x>3$$

The final solution is the union of the obtained solutions.

$$x<-1\text{; }x>3$$