Solve (9-5x)^2+2(9-5x)^2-24<0

$$( 9 - 5 x ) ^ { 2 } + 2 ( 9 - 5 x ) ^ { 2 } - 24 < 0$$

Solve for x

$$x\in \left(\frac{9-2\sqrt{2}}{5},\frac{2\sqrt{2}+9}{5}\right)$$

Steps Using the Quadratic Formula

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

$$81-90x+25x^{2}+2\left(9-5x\right)^{2}-24<0$$

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

$$81-90x+25x^{2}+2\left(81-90x+25x^{2}\right)-24<0$$

Use the distributive property to multiply 2 by 81-90x+25x^{2}.

$$81-90x+25x^{2}+162-180x+50x^{2}-24<0$$

Add 81 and 162 to get 243.

$$243-90x+25x^{2}-180x+50x^{2}-24<0$$

Combine -90x and -180x to get -270x.

$$243-270x+25x^{2}+50x^{2}-24<0$$

Combine 25x^{2} and 50x^{2} to get 75x^{2}.

$$243-270x+75x^{2}-24<0$$

Subtract 24 from 243 to get 219.

$$219-270x+75x^{2}<0$$

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.

$$219-270x+75x^{2}=0$$

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 75 for a, -270 for b, and 219 for c in the quadratic formula.

$$x=\frac{-\left(-270\right)±\sqrt{\left(-270\right)^{2}-4\times 75\times 219}}{2\times 75}$$

Do the calculations.

$$x=\frac{270±60\sqrt{2}}{150}$$

Solve the equation x=\frac{270±60\sqrt{2}}{150} when ± is plus and when ± is minus.

$$x=\frac{2\sqrt{2}+9}{5} x=\frac{9-2\sqrt{2}}{5}$$

Rewrite the inequality by using the obtained solutions.

$$75\left(x-\frac{2\sqrt{2}+9}{5}\right)\left(x-\frac{9-2\sqrt{2}}{5}\right)<0$$

For the product to be negative, x-\frac{2\sqrt{2}+9}{5} and x-\frac{9-2\sqrt{2}}{5} have to be of the opposite signs. Consider the case when x-\frac{2\sqrt{2}+9}{5} is positive and x-\frac{9-2\sqrt{2}}{5} is negative.

$$x-\frac{2\sqrt{2}+9}{5}>0 x-\frac{9-2\sqrt{2}}{5}<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\frac{9-2\sqrt{2}}{5} is positive and x-\frac{2\sqrt{2}+9}{5} is negative.

$$x-\frac{9-2\sqrt{2}}{5}>0 x-\frac{2\sqrt{2}+9}{5}<0$$

The solution satisfying both inequalities is x\in \left(\frac{9-2\sqrt{2}}{5},\frac{2\sqrt{2}+9}{5}\right).

$$x\in \left(\frac{9-2\sqrt{2}}{5},\frac{2\sqrt{2}+9}{5}\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(\frac{9-2\sqrt{2}}{5},\frac{2\sqrt{2}+9}{5}\right)$$