Solve (4x^2)-4(1)(-4x+8)<0

$$(4 { x }^{ 2 } )-4(1)(-4x+8) < 0$$

Solve for x

$$x\in \left(-2\sqrt{3}-2,2\sqrt{3}-2\right)$$

Steps Using the Quadratic Formula

Multiply 4 and 1 to get 4.

$$4x^{2}-4\left(-4x+8\right)<0$$

Use the distributive property to multiply -4 by -4x+8.

$$4x^{2}+16x-32<0$$

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.

$$4x^{2}+16x-32=0$$

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 4 for a, 16 for b, and -32 for c in the quadratic formula.

$$x=\frac{-16±\sqrt{16^{2}-4\times 4\left(-32\right)}}{2\times 4}$$

Do the calculations.

$$x=\frac{-16±16\sqrt{3}}{8}$$

Solve the equation x=\frac{-16±16\sqrt{3}}{8} when ± is plus and when ± is minus.

$$x=2\sqrt{3}-2 x=-2\sqrt{3}-2$$

Rewrite the inequality by using the obtained solutions.

$$4\left(x-\left(2\sqrt{3}-2\right)\right)\left(x-\left(-2\sqrt{3}-2\right)\right)<0$$

For the product to be negative, x-\left(2\sqrt{3}-2\right) and x-\left(-2\sqrt{3}-2\right) have to be of the opposite signs. Consider the case when x-\left(2\sqrt{3}-2\right) is positive and x-\left(-2\sqrt{3}-2\right) is negative.

$$x-\left(2\sqrt{3}-2\right)>0 x-\left(-2\sqrt{3}-2\right)<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\left(-2\sqrt{3}-2\right) is positive and x-\left(2\sqrt{3}-2\right) is negative.

$$x-\left(-2\sqrt{3}-2\right)>0 x-\left(2\sqrt{3}-2\right)<0$$

The solution satisfying both inequalities is x\in \left(-2\sqrt{3}-2,2\sqrt{3}-2\right).

$$x\in \left(-2\sqrt{3}-2,2\sqrt{3}-2\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(-2\sqrt{3}-2,2\sqrt{3}-2\right)$$