Solve 4n^2+8n-1292=0

$$4 { n }^{ 2 } +8n-1292=0$$

Solve for n

$$n=-19$$

$$n=17$$

Steps Using Factoring By Grouping

Divide both sides by 4.

$$n^{2}+2n-323=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as n^{2}+an+bn-323. To find a and b, set up a system to be solved.

$$a+b=2 ab=1\left(-323\right)=-323$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -323.

$$-1,323 -17,19$$

Calculate the sum for each pair.

$$-1+323=322 -17+19=2$$

The solution is the pair that gives sum 2.

$$a=-17 b=19$$

Rewrite n^{2}+2n-323 as \left(n^{2}-17n\right)+\left(19n-323\right).

$$\left(n^{2}-17n\right)+\left(19n-323\right)$$

Factor out n in the first and 19 in the second group.

$$n\left(n-17\right)+19\left(n-17\right)$$

Factor out common term n-17 by using distributive property.

$$\left(n-17\right)\left(n+19\right)$$

To find equation solutions, solve n-17=0 and n+19=0.

$$n=17 n=-19$$

Steps Using the Quadratic Formula

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$4n^{2}+8n-1292=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 8 for b, and -1292 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-8±\sqrt{8^{2}-4\times 4\left(-1292\right)}}{2\times 4}$$

Square 8.

$$n=\frac{-8±\sqrt{64-4\times 4\left(-1292\right)}}{2\times 4}$$

Multiply -4 times 4.

$$n=\frac{-8±\sqrt{64-16\left(-1292\right)}}{2\times 4}$$

Multiply -16 times -1292.

$$n=\frac{-8±\sqrt{64+20672}}{2\times 4}$$

Add 64 to 20672.

$$n=\frac{-8±\sqrt{20736}}{2\times 4}$$

Take the square root of 20736.

$$n=\frac{-8±144}{2\times 4}$$

Multiply 2 times 4.

$$n=\frac{-8±144}{8}$$

Now solve the equation n=\frac{-8±144}{8} when ± is plus. Add -8 to 144.

$$n=\frac{136}{8}$$

Divide 136 by 8.

$$n=17$$

Now solve the equation n=\frac{-8±144}{8} when ± is minus. Subtract 144 from -8.

$$n=-\frac{152}{8}$$

Divide -152 by 8.

$$n=-19$$

The equation is now solved.

$$n=17 n=-19$$

Steps for Completing the Square

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$4n^{2}+8n-1292=0$$

Add 1292 to both sides of the equation.

$$4n^{2}+8n-1292-\left(-1292\right)=-\left(-1292\right)$$

Subtracting -1292 from itself leaves 0.

$$4n^{2}+8n=-\left(-1292\right)$$

Subtract -1292 from 0.

$$4n^{2}+8n=1292$$

Divide both sides by 4.

$$\frac{4n^{2}+8n}{4}=\frac{1292}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$n^{2}+\frac{8}{4}n=\frac{1292}{4}$$

Divide 8 by 4.

$$n^{2}+2n=\frac{1292}{4}$$

Divide 1292 by 4.

$$n^{2}+2n=323$$

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}+2n+1^{2}=323+1^{2}$$

Square 1.

$$n^{2}+2n+1=323+1$$

Add 323 to 1.

$$n^{2}+2n+1=324$$

Factor n^{2}+2n+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n+1\right)^{2}=324$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n+1\right)^{2}}=\sqrt{324}$$

Simplify.

$$n+1=18 n+1=-18$$

Subtract 1 from both sides of the equation.

$$n=17 n=-19$$