Solve (9-5a)^2-4*4a^2<0

$$( 9 - 5 a ) ^ { 2 } - 4 \times 4 a ^ { 2 } < 0$$

Solve for a

$$a\in \left(1,9\right)$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(9-5a\right)^{2}.

$$81-90a+25a^{2}-4\times 4a^{2}<0$$

Multiply 4 and 4 to get 16.

$$81-90a+25a^{2}-16a^{2}<0$$

Combine 25a^{2} and -16a^{2} to get 9a^{2}.

$$81-90a+9a^{2}<0$$

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.

$$81-90a+9a^{2}=0$$

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 9 for a, -90 for b, and 81 for c in the quadratic formula.

$$a=\frac{-\left(-90\right)±\sqrt{\left(-90\right)^{2}-4\times 9\times 81}}{2\times 9}$$

Do the calculations.

$$a=\frac{90±72}{18}$$

Solve the equation a=\frac{90±72}{18} when ± is plus and when ± is minus.

$$a=9 a=1$$

Rewrite the inequality by using the obtained solutions.

$$9\left(a-9\right)\left(a-1\right)<0$$

For the product to be negative, a-9 and a-1 have to be of the opposite signs. Consider the case when a-9 is positive and a-1 is negative.

$$a-9>0 a-1<0$$

This is false for any a.

$$a\in \emptyset$$

Consider the case when a-1 is positive and a-9 is negative.

$$a-1>0 a-9<0$$

The solution satisfying both inequalities is a\in \left(1,9\right).

$$a\in \left(1,9\right)$$

The final solution is the union of the obtained solutions.

$$a\in \left(1,9\right)$$