Solve 4a^2-24a(a-3)>0

$$4 a ^ { 2 } - 24 a ( a - 3 ) > 0$$

Solve for a

$$a\in \left(0,\frac{18}{5}\right)$$

Solution Steps

Multiply -1 and 24 to get -24.

$$4a^{2}-24a\left(a-3\right)>0$$

Use the distributive property to multiply -24a by a-3.

$$4a^{2}-24a^{2}+72a>0$$

Combine 4a^{2} and -24a^{2} to get -20a^{2}.

$$-20a^{2}+72a>0$$

Multiply the inequality by -1 to make the coefficient of the highest power in -20a^{2}+72a positive. Since -1 is negative, the inequality direction is changed.

$$20a^{2}-72a<0$$

Factor out a.

$$4a\left(5a-18\right)<0$$

For the product to be negative, a and a-\frac{18}{5} have to be of the opposite signs. Consider the case when a is positive and a-\frac{18}{5} is negative.

$$a>0 a-\frac{18}{5}<0$$

The solution satisfying both inequalities is a\in \left(0,\frac{18}{5}\right).

$$a\in \left(0,\frac{18}{5}\right)$$

Consider the case when a-\frac{18}{5} is positive and a is negative.

$$a-\frac{18}{5}>0 a<0$$

This is false for any a.

$$a\in \emptyset$$

The final solution is the union of the obtained solutions.

$$a\in \left(0,\frac{18}{5}\right)$$