Solve (m+1)^2-4(m^2-4)<0

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

Solve for m

$$m\in \left(-\infty,\frac{1-2\sqrt{13}}{3}\right)\cup \left(\frac{2\sqrt{13}+1}{3},\infty\right)$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(m+1\right)^{2}.

$$m^{2}+2m+1-4\left(m^{2}-4\right)<0$$

Use the distributive property to multiply -4 by m^{2}-4.

$$m^{2}+2m+1-4m^{2}+16<0$$

Combine m^{2} and -4m^{2} to get -3m^{2}.

$$-3m^{2}+2m+1+16<0$$

Add 1 and 16 to get 17.

$$-3m^{2}+2m+17<0$$

Multiply the inequality by -1 to make the coefficient of the highest power in -3m^{2}+2m+17 positive. Since -1 is negative, the inequality direction is changed.

$$3m^{2}-2m-17>0$$

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.

$$3m^{2}-2m-17=0$$

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 3 for a, -2 for b, and -17 for c in the quadratic formula.

$$m=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 3\left(-17\right)}}{2\times 3}$$

Do the calculations.

$$m=\frac{2±4\sqrt{13}}{6}$$

Solve the equation m=\frac{2±4\sqrt{13}}{6} when ± is plus and when ± is minus.

$$m=\frac{2\sqrt{13}+1}{3} m=\frac{1-2\sqrt{13}}{3}$$

Rewrite the inequality by using the obtained solutions.

$$3\left(m-\frac{2\sqrt{13}+1}{3}\right)\left(m-\frac{1-2\sqrt{13}}{3}\right)>0$$

For the product to be positive, m-\frac{2\sqrt{13}+1}{3} and m-\frac{1-2\sqrt{13}}{3} have to be both negative or both positive. Consider the case when m-\frac{2\sqrt{13}+1}{3} and m-\frac{1-2\sqrt{13}}{3} are both negative.

$$m-\frac{2\sqrt{13}+1}{3}<0 m-\frac{1-2\sqrt{13}}{3}<0$$

The solution satisfying both inequalities is m<\frac{1-2\sqrt{13}}{3}.

$$m<\frac{1-2\sqrt{13}}{3}$$

Consider the case when m-\frac{2\sqrt{13}+1}{3} and m-\frac{1-2\sqrt{13}}{3} are both positive.

$$m-\frac{1-2\sqrt{13}}{3}>0 m-\frac{2\sqrt{13}+1}{3}>0$$

The solution satisfying both inequalities is m>\frac{2\sqrt{13}+1}{3}.

$$m>\frac{2\sqrt{13}+1}{3}$$

The final solution is the union of the obtained solutions.

$$m<\frac{1-2\sqrt{13}}{3}\text{; }m>\frac{2\sqrt{13}+1}{3}$$