Solve 16-(-a)(-a+3)4<0

$$16 - ( - a ) ( - a + 3 ) 4 < 0$$

Solve for a

$$a\in \left(-\infty,-1\right)\cup \left(4,\infty\right)$$

Steps Using the Quadratic Formula

Use the distributive property to multiply -a by -a+3.

$$16-\left(\left(-a\right)^{2}+3\left(-a\right)\right)\times 4<0$$

Calculate -a to the power of 2 and get a^{2}.

$$16-\left(a^{2}+3\left(-a\right)\right)\times 4<0$$

Use the distributive property to multiply a^{2}+3\left(-a\right) by 4.

$$16-\left(4a^{2}+12\left(-a\right)\right)<0$$

To find the opposite of 4a^{2}+12\left(-a\right), find the opposite of each term.

$$16-4a^{2}-12\left(-a\right)<0$$

Multiply -12 and -1 to get 12.

$$16-4a^{2}+12a<0$$

Multiply the inequality by -1 to make the coefficient of the highest power in 16-4a^{2}+12a positive. Since -1 is negative, the inequality direction is changed.

$$-16+4a^{2}-12a>0$$

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.

$$-16+4a^{2}-12a=0$$

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 4 for a, -12 for b, and -16 for c in the quadratic formula.

$$a=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 4\left(-16\right)}}{2\times 4}$$

Do the calculations.

$$a=\frac{12±20}{8}$$

Solve the equation a=\frac{12±20}{8} when ± is plus and when ± is minus.

$$a=4 a=-1$$

Rewrite the inequality by using the obtained solutions.

$$4\left(a-4\right)\left(a+1\right)>0$$

For the product to be positive, a-4 and a+1 have to be both negative or both positive. Consider the case when a-4 and a+1 are both negative.

$$a-4<0 a+1<0$$

The solution satisfying both inequalities is a<-1.

$$a<-1$$

Consider the case when a-4 and a+1 are both positive.

$$a+1>0 a-4>0$$

The solution satisfying both inequalities is a>4.

$$a>4$$

The final solution is the union of the obtained solutions.

$$a<-1\text{; }a>4$$