Solve left(4x-12right)^2-4x(15-x)geq0

$${ \left(4x-12 \right) }^{ 2 } -4x(15-x) \geq 0$$

Solve for x

$$x\in (-\infty,\frac{39-3\sqrt{89}}{10}]\cup [\frac{3\sqrt{89}+39}{10},\infty)$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4x-12\right)^{2}.

$$16x^{2}-96x+144-4x\left(15-x\right)\geq 0$$

Use the distributive property to multiply -4x by 15-x.

$$16x^{2}-96x+144-60x+4x^{2}\geq 0$$

Combine -96x and -60x to get -156x.

$$16x^{2}-156x+144+4x^{2}\geq 0$$

Combine 16x^{2} and 4x^{2} to get 20x^{2}.

$$20x^{2}-156x+144\geq 0$$

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.

$$20x^{2}-156x+144=0$$

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 20 for a, -156 for b, and 144 for c in the quadratic formula.

$$x=\frac{-\left(-156\right)±\sqrt{\left(-156\right)^{2}-4\times 20\times 144}}{2\times 20}$$

Do the calculations.

$$x=\frac{156±12\sqrt{89}}{40}$$

Solve the equation x=\frac{156±12\sqrt{89}}{40} when ± is plus and when ± is minus.

$$x=\frac{3\sqrt{89}+39}{10} x=\frac{39-3\sqrt{89}}{10}$$

Rewrite the inequality by using the obtained solutions.

$$20\left(x-\frac{3\sqrt{89}+39}{10}\right)\left(x-\frac{39-3\sqrt{89}}{10}\right)\geq 0$$

For the product to be ≥0, x-\frac{3\sqrt{89}+39}{10} and x-\frac{39-3\sqrt{89}}{10} have to be both ≤0 or both ≥0. Consider the case when x-\frac{3\sqrt{89}+39}{10} and x-\frac{39-3\sqrt{89}}{10} are both ≤0.

$$x-\frac{3\sqrt{89}+39}{10}\leq 0 x-\frac{39-3\sqrt{89}}{10}\leq 0$$

The solution satisfying both inequalities is x\leq \frac{39-3\sqrt{89}}{10}.

$$x\leq \frac{39-3\sqrt{89}}{10}$$

Consider the case when x-\frac{3\sqrt{89}+39}{10} and x-\frac{39-3\sqrt{89}}{10} are both ≥0.

$$x-\frac{39-3\sqrt{89}}{10}\geq 0 x-\frac{3\sqrt{89}+39}{10}\geq 0$$

The solution satisfying both inequalities is x\geq \frac{3\sqrt{89}+39}{10}.

$$x\geq \frac{3\sqrt{89}+39}{10}$$

The final solution is the union of the obtained solutions.

$$x\leq \frac{39-3\sqrt{89}}{10}\text{; }x\geq \frac{3\sqrt{89}+39}{10}$$