Solve 80(n-2)=n[5(n-2)]

$$80 ( n - 2 ) = n [ 5 ( n - 2 ) ]$$

Solve for n

$$n=16$$

$$n=2$$

Steps Using Factoring By Grouping

Use the distributive property to multiply 80 by n-2.

$$80n-160=n\times 5\left(n-2\right)$$

Use the distributive property to multiply n\times 5 by n-2.

$$80n-160=5n^{2}-2n\times 5$$

Multiply -2 and 5 to get -10.

$$80n-160=5n^{2}-10n$$

Subtract 5n^{2} from both sides.

$$80n-160-5n^{2}=-10n$$

Add 10n to both sides.

$$80n-160-5n^{2}+10n=0$$

Combine 80n and 10n to get 90n.

$$90n-160-5n^{2}=0$$

Divide both sides by 5.

$$18n-32-n^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-n^{2}+18n-32=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -n^{2}+an+bn-32. To find a and b, set up a system to be solved.

$$a+b=18 ab=-\left(-32\right)=32$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 32.

$$1,32 2,16 4,8$$

Calculate the sum for each pair.

$$1+32=33 2+16=18 4+8=12$$

The solution is the pair that gives sum 18.

$$a=16 b=2$$

Rewrite -n^{2}+18n-32 as \left(-n^{2}+16n\right)+\left(2n-32\right).

$$\left(-n^{2}+16n\right)+\left(2n-32\right)$$

Factor out -n in the first and 2 in the second group.

$$-n\left(n-16\right)+2\left(n-16\right)$$

Factor out common term n-16 by using distributive property.

$$\left(n-16\right)\left(-n+2\right)$$

To find equation solutions, solve n-16=0 and -n+2=0.

$$n=16 n=2$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 80 by n-2.

$$80n-160=n\times 5\left(n-2\right)$$

Use the distributive property to multiply n\times 5 by n-2.

$$80n-160=5n^{2}-2n\times 5$$

Multiply -2 and 5 to get -10.

$$80n-160=5n^{2}-10n$$

Subtract 5n^{2} from both sides.

$$80n-160-5n^{2}=-10n$$

Add 10n to both sides.

$$80n-160-5n^{2}+10n=0$$

Combine 80n and 10n to get 90n.

$$90n-160-5n^{2}=0$$

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

$$-5n^{2}+90n-160=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 90 for b, and -160 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-90±\sqrt{90^{2}-4\left(-5\right)\left(-160\right)}}{2\left(-5\right)}$$

Square 90.

$$n=\frac{-90±\sqrt{8100-4\left(-5\right)\left(-160\right)}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$n=\frac{-90±\sqrt{8100+20\left(-160\right)}}{2\left(-5\right)}$$

Multiply 20 times -160.

$$n=\frac{-90±\sqrt{8100-3200}}{2\left(-5\right)}$$

Add 8100 to -3200.

$$n=\frac{-90±\sqrt{4900}}{2\left(-5\right)}$$

Take the square root of 4900.

$$n=\frac{-90±70}{2\left(-5\right)}$$

Multiply 2 times -5.

$$n=\frac{-90±70}{-10}$$

Now solve the equation n=\frac{-90±70}{-10} when ± is plus. Add -90 to 70.

$$n=-\frac{20}{-10}$$

Divide -20 by -10.

$$n=2$$

Now solve the equation n=\frac{-90±70}{-10} when ± is minus. Subtract 70 from -90.

$$n=-\frac{160}{-10}$$

Divide -160 by -10.

$$n=16$$

The equation is now solved.

$$n=2 n=16$$

Steps for Completing the Square

Use the distributive property to multiply 80 by n-2.

$$80n-160=n\times 5\left(n-2\right)$$

Use the distributive property to multiply n\times 5 by n-2.

$$80n-160=5n^{2}-2n\times 5$$

Multiply -2 and 5 to get -10.

$$80n-160=5n^{2}-10n$$

Subtract 5n^{2} from both sides.

$$80n-160-5n^{2}=-10n$$

Add 10n to both sides.

$$80n-160-5n^{2}+10n=0$$

Combine 80n and 10n to get 90n.

$$90n-160-5n^{2}=0$$

Add 160 to both sides. Anything plus zero gives itself.

$$90n-5n^{2}=160$$

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.

$$-5n^{2}+90n=160$$

Divide both sides by -5.

$$\frac{-5n^{2}+90n}{-5}=\frac{160}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$n^{2}+\frac{90}{-5}n=\frac{160}{-5}$$

Divide 90 by -5.

$$n^{2}-18n=\frac{160}{-5}$$

Divide 160 by -5.

$$n^{2}-18n=-32$$

Divide -18, the coefficient of the x term, by 2 to get -9. Then add the square of -9 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}-18n+\left(-9\right)^{2}=-32+\left(-9\right)^{2}$$

Square -9.

$$n^{2}-18n+81=-32+81$$

Add -32 to 81.

$$n^{2}-18n+81=49$$

Factor n^{2}-18n+81. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n-9\right)^{2}=49$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-9\right)^{2}}=\sqrt{49}$$

Simplify.

$$n-9=7 n-9=-7$$

Add 9 to both sides of the equation.

$$n=16 n=2$$