Solve (2k+5)^2-40>0

$$( 2 k + 5 ) ^ { 2 } - 40 > 0$$

Solve for k

$$k\in \left(-\infty,-\sqrt{10}-\frac{5}{2}\right)\cup \left(\sqrt{10}-\frac{5}{2},\infty\right)$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2k+5\right)^{2}.

$$4k^{2}+20k+25-40>0$$

Subtract 40 from 25 to get -15.

$$4k^{2}+20k-15>0$$

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.

$$4k^{2}+20k-15=0$$

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, 20 for b, and -15 for c in the quadratic formula.

$$k=\frac{-20±\sqrt{20^{2}-4\times 4\left(-15\right)}}{2\times 4}$$

Do the calculations.

$$k=\frac{-20±8\sqrt{10}}{8}$$

Solve the equation k=\frac{-20±8\sqrt{10}}{8} when ± is plus and when ± is minus.

$$k=\sqrt{10}-\frac{5}{2} k=-\sqrt{10}-\frac{5}{2}$$

Rewrite the inequality by using the obtained solutions.

$$4\left(k-\left(\sqrt{10}-\frac{5}{2}\right)\right)\left(k-\left(-\sqrt{10}-\frac{5}{2}\right)\right)>0$$

For the product to be positive, k-\left(\sqrt{10}-\frac{5}{2}\right) and k-\left(-\sqrt{10}-\frac{5}{2}\right) have to be both negative or both positive. Consider the case when k-\left(\sqrt{10}-\frac{5}{2}\right) and k-\left(-\sqrt{10}-\frac{5}{2}\right) are both negative.

$$k-\left(\sqrt{10}-\frac{5}{2}\right)<0 k-\left(-\sqrt{10}-\frac{5}{2}\right)<0$$

The solution satisfying both inequalities is k<-\sqrt{10}-\frac{5}{2}.

$$k<-\sqrt{10}-\frac{5}{2}$$

Consider the case when k-\left(\sqrt{10}-\frac{5}{2}\right) and k-\left(-\sqrt{10}-\frac{5}{2}\right) are both positive.

$$k-\left(-\sqrt{10}-\frac{5}{2}\right)>0 k-\left(\sqrt{10}-\frac{5}{2}\right)>0$$

The solution satisfying both inequalities is k>\sqrt{10}-\frac{5}{2}.

$$k>\sqrt{10}-\frac{5}{2}$$

The final solution is the union of the obtained solutions.

$$k<-\sqrt{10}-\frac{5}{2}\text{; }k>\sqrt{10}-\frac{5}{2}$$