Solve 12-x(x-4)<0

$$12 - x ( x - 4 ) < 0$$

Solve for x

$$x\in \left(-\infty,-2\right)\cup \left(6,\infty\right)$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x by x-4.

$$12-\left(x^{2}-4x\right)<0$$

To find the opposite of x^{2}-4x, find the opposite of each term.

$$12-x^{2}-\left(-4x\right)<0$$

The opposite of -4x is 4x.

$$12-x^{2}+4x<0$$

Multiply the inequality by -1 to make the coefficient of the highest power in 12-x^{2}+4x positive. Since -1 is negative, the inequality direction is changed.

$$-12+x^{2}-4x>0$$

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.

$$-12+x^{2}-4x=0$$

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, -4 for b, and -12 for c in the quadratic formula.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 1\left(-12\right)}}{2}$$

Do the calculations.

$$x=\frac{4±8}{2}$$

Solve the equation x=\frac{4±8}{2} when ± is plus and when ± is minus.

$$x=6 x=-2$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-6\right)\left(x+2\right)>0$$

For the product to be positive, x-6 and x+2 have to be both negative or both positive. Consider the case when x-6 and x+2 are both negative.

$$x-6<0 x+2<0$$

The solution satisfying both inequalities is x<-2.

$$x<-2$$

Consider the case when x-6 and x+2 are both positive.

$$x+2>0 x-6>0$$

The solution satisfying both inequalities is x>6.

$$x>6$$

The final solution is the union of the obtained solutions.

$$x<-2\text{; }x>6$$