Solve -3(1-x)(x-3)>0

$$- 3 ( 1 - x ) ( x - 3 ) > 0$$

Solve for x

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

Steps Using the Quadratic Formula

Use the distributive property to multiply -3 by 1-x.

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

Use the distributive property to multiply -3+3x by x-3 and combine like terms.

$$-12x+9+3x^{2}>0$$

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.

$$-12x+9+3x^{2}=0$$

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 3 for a, -12 for b, and 9 for c in the quadratic formula.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 3\times 9}}{2\times 3}$$

Do the calculations.

$$x=\frac{12±6}{6}$$

Solve the equation x=\frac{12±6}{6} when ± is plus and when ± is minus.

$$x=3 x=1$$

Rewrite the inequality by using the obtained solutions.

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

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.

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