Solve 0>4n^2+2n-1600

$$0 > 4 { n }^{ 2 } +2n-1600$$

Solve for n

$$n\in \left(\frac{-\sqrt{6401}-1}{4},\frac{\sqrt{6401}-1}{4}\right)$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side. This changes the sign direction.

$$4n^{2}+2n-1600<0$$

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.

$$4n^{2}+2n-1600=0$$

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, 2 for b, and -1600 for c in the quadratic formula.

$$n=\frac{-2±\sqrt{2^{2}-4\times 4\left(-1600\right)}}{2\times 4}$$

Do the calculations.

$$n=\frac{-2±2\sqrt{6401}}{8}$$

Solve the equation n=\frac{-2±2\sqrt{6401}}{8} when ± is plus and when ± is minus.

$$n=\frac{\sqrt{6401}-1}{4} n=\frac{-\sqrt{6401}-1}{4}$$

Rewrite the inequality by using the obtained solutions.

$$4\left(n-\frac{\sqrt{6401}-1}{4}\right)\left(n-\frac{-\sqrt{6401}-1}{4}\right)<0$$

For the product to be negative, n-\frac{\sqrt{6401}-1}{4} and n-\frac{-\sqrt{6401}-1}{4} have to be of the opposite signs. Consider the case when n-\frac{\sqrt{6401}-1}{4} is positive and n-\frac{-\sqrt{6401}-1}{4} is negative.

$$n-\frac{\sqrt{6401}-1}{4}>0 n-\frac{-\sqrt{6401}-1}{4}<0$$

This is false for any n.

$$n\in \emptyset$$

Consider the case when n-\frac{-\sqrt{6401}-1}{4} is positive and n-\frac{\sqrt{6401}-1}{4} is negative.

$$n-\frac{-\sqrt{6401}-1}{4}>0 n-\frac{\sqrt{6401}-1}{4}<0$$

The solution satisfying both inequalities is n\in \left(\frac{-\sqrt{6401}-1}{4},\frac{\sqrt{6401}-1}{4}\right).

$$n\in \left(\frac{-\sqrt{6401}-1}{4},\frac{\sqrt{6401}-1}{4}\right)$$

The final solution is the union of the obtained solutions.

$$n\in \left(\frac{-\sqrt{6401}-1}{4},\frac{\sqrt{6401}-1}{4}\right)$$