Solve left(6x-2right)^2-4*x*5x<0

$${ \left(6x-2 \right) }^{ 2 } -4 \times x \times 5x < 0$$

Solve for x

$$x\in \left(\frac{3-\sqrt{5}}{4},\frac{\sqrt{5}+3}{4}\right)$$

Steps Using the Quadratic Formula

Multiply x and x to get x^{2}.

$$\left(6x-2\right)^{2}-4x^{2}\times 5<0$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(6x-2\right)^{2}.

$$36x^{2}-24x+4-4x^{2}\times 5<0$$

Multiply 4 and 5 to get 20.

$$36x^{2}-24x+4-20x^{2}<0$$

Combine 36x^{2} and -20x^{2} to get 16x^{2}.

$$16x^{2}-24x+4<0$$

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.

$$16x^{2}-24x+4=0$$

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 16 for a, -24 for b, and 4 for c in the quadratic formula.

$$x=\frac{-\left(-24\right)±\sqrt{\left(-24\right)^{2}-4\times 16\times 4}}{2\times 16}$$

Do the calculations.

$$x=\frac{24±8\sqrt{5}}{32}$$

Solve the equation x=\frac{24±8\sqrt{5}}{32} when ± is plus and when ± is minus.

$$x=\frac{\sqrt{5}+3}{4} x=\frac{3-\sqrt{5}}{4}$$

Rewrite the inequality by using the obtained solutions.

$$16\left(x-\frac{\sqrt{5}+3}{4}\right)\left(x-\frac{3-\sqrt{5}}{4}\right)<0$$

For the product to be negative, x-\frac{\sqrt{5}+3}{4} and x-\frac{3-\sqrt{5}}{4} have to be of the opposite signs. Consider the case when x-\frac{\sqrt{5}+3}{4} is positive and x-\frac{3-\sqrt{5}}{4} is negative.

$$x-\frac{\sqrt{5}+3}{4}>0 x-\frac{3-\sqrt{5}}{4}<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\frac{3-\sqrt{5}}{4} is positive and x-\frac{\sqrt{5}+3}{4} is negative.

$$x-\frac{3-\sqrt{5}}{4}>0 x-\frac{\sqrt{5}+3}{4}<0$$

The solution satisfying both inequalities is x\in \left(\frac{3-\sqrt{5}}{4},\frac{\sqrt{5}+3}{4}\right).

$$x\in \left(\frac{3-\sqrt{5}}{4},\frac{\sqrt{5}+3}{4}\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(\frac{3-\sqrt{5}}{4},\frac{\sqrt{5}+3}{4}\right)$$