Solve left(4xright)^2-4(5)(x^2-2x)>0

$${ \left(4x \right) }^{ 2 } -4(5)( { x }^{ 2 } -2x) > 0$$

Solve for x

$$x\in \left(0,10\right)$$

Solution Steps

Expand \left(4x\right)^{2}.

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

Calculate 4 to the power of 2 and get 16.

$$16x^{2}-4\times 5\left(x^{2}-2x\right)>0$$

Multiply 4 and 5 to get 20.

$$16x^{2}-20\left(x^{2}-2x\right)>0$$

Use the distributive property to multiply -20 by x^{2}-2x.

$$16x^{2}-20x^{2}+40x>0$$

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

$$-4x^{2}+40x>0$$

Multiply the inequality by -1 to make the coefficient of the highest power in -4x^{2}+40x positive. Since -1 is negative, the inequality direction is changed.

$$4x^{2}-40x<0$$

Factor out x.

$$4x\left(x-10\right)<0$$

For the product to be negative, x and x-10 have to be of the opposite signs. Consider the case when x is positive and x-10 is negative.

$$x>0 x-10<0$$

The solution satisfying both inequalities is x\in \left(0,10\right).

$$x\in \left(0,10\right)$$

Consider the case when x-10 is positive and x is negative.

$$x-10>0 x<0$$

This is false for any x.

$$x\in \emptyset$$

The final solution is the union of the obtained solutions.

$$x\in \left(0,10\right)$$