Solve 12(n-4)-30=n^2-9n+12

$$12 ( n - 4 ) - 30 = n ^ { 2 } - 9 n + 12$$

Solve for n

$$n=6$$

$$n=15$$

Steps Using Factoring By Grouping

Use the distributive property to multiply 12 by n-4.

$$12n-48-30=n^{2}-9n+12$$

Subtract 30 from -48 to get -78.

$$12n-78=n^{2}-9n+12$$

Subtract n^{2} from both sides.

$$12n-78-n^{2}=-9n+12$$

Add 9n to both sides.

$$12n-78-n^{2}+9n=12$$

Combine 12n and 9n to get 21n.

$$21n-78-n^{2}=12$$

Subtract 12 from both sides.

$$21n-78-n^{2}-12=0$$

Subtract 12 from -78 to get -90.

$$21n-90-n^{2}=0$$

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

$$-n^{2}+21n-90=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -n^{2}+an+bn-90. To find a and b, set up a system to be solved.

$$a+b=21 ab=-\left(-90\right)=90$$

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 90.

$$1,90 2,45 3,30 5,18 6,15 9,10$$

Calculate the sum for each pair.

$$1+90=91 2+45=47 3+30=33 5+18=23 6+15=21 9+10=19$$

The solution is the pair that gives sum 21.

$$a=15 b=6$$

Rewrite -n^{2}+21n-90 as \left(-n^{2}+15n\right)+\left(6n-90\right).

$$\left(-n^{2}+15n\right)+\left(6n-90\right)$$

Factor out -n in the first and 6 in the second group.

$$-n\left(n-15\right)+6\left(n-15\right)$$

Factor out common term n-15 by using distributive property.

$$\left(n-15\right)\left(-n+6\right)$$

To find equation solutions, solve n-15=0 and -n+6=0.

$$n=15 n=6$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 12 by n-4.

$$12n-48-30=n^{2}-9n+12$$

Subtract 30 from -48 to get -78.

$$12n-78=n^{2}-9n+12$$

Subtract n^{2} from both sides.

$$12n-78-n^{2}=-9n+12$$

Add 9n to both sides.

$$12n-78-n^{2}+9n=12$$

Combine 12n and 9n to get 21n.

$$21n-78-n^{2}=12$$

Subtract 12 from both sides.

$$21n-78-n^{2}-12=0$$

Subtract 12 from -78 to get -90.

$$21n-90-n^{2}=0$$

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.

$$-n^{2}+21n-90=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 21 for b, and -90 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-21±\sqrt{21^{2}-4\left(-1\right)\left(-90\right)}}{2\left(-1\right)}$$

Square 21.

$$n=\frac{-21±\sqrt{441-4\left(-1\right)\left(-90\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$n=\frac{-21±\sqrt{441+4\left(-90\right)}}{2\left(-1\right)}$$

Multiply 4 times -90.

$$n=\frac{-21±\sqrt{441-360}}{2\left(-1\right)}$$

Add 441 to -360.

$$n=\frac{-21±\sqrt{81}}{2\left(-1\right)}$$

Take the square root of 81.

$$n=\frac{-21±9}{2\left(-1\right)}$$

Multiply 2 times -1.

$$n=\frac{-21±9}{-2}$$

Now solve the equation n=\frac{-21±9}{-2} when ± is plus. Add -21 to 9.

$$n=-\frac{12}{-2}$$

Divide -12 by -2.

$$n=6$$

Now solve the equation n=\frac{-21±9}{-2} when ± is minus. Subtract 9 from -21.

$$n=-\frac{30}{-2}$$

Divide -30 by -2.

$$n=15$$

The equation is now solved.

$$n=6 n=15$$

Steps for Completing the Square

Use the distributive property to multiply 12 by n-4.

$$12n-48-30=n^{2}-9n+12$$

Subtract 30 from -48 to get -78.

$$12n-78=n^{2}-9n+12$$

Subtract n^{2} from both sides.

$$12n-78-n^{2}=-9n+12$$

Add 9n to both sides.

$$12n-78-n^{2}+9n=12$$

Combine 12n and 9n to get 21n.

$$21n-78-n^{2}=12$$

Add 78 to both sides.

$$21n-n^{2}=12+78$$

Add 12 and 78 to get 90.

$$21n-n^{2}=90$$

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.

$$-n^{2}+21n=90$$

Divide both sides by -1.

$$\frac{-n^{2}+21n}{-1}=\frac{90}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$n^{2}+\frac{21}{-1}n=\frac{90}{-1}$$

Divide 21 by -1.

$$n^{2}-21n=\frac{90}{-1}$$

Divide 90 by -1.

$$n^{2}-21n=-90$$

Divide -21, the coefficient of the x term, by 2 to get -\frac{21}{2}. Then add the square of -\frac{21}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}-21n+\left(-\frac{21}{2}\right)^{2}=-90+\left(-\frac{21}{2}\right)^{2}$$

Square -\frac{21}{2} by squaring both the numerator and the denominator of the fraction.

$$n^{2}-21n+\frac{441}{4}=-90+\frac{441}{4}$$

Add -90 to \frac{441}{4}.

$$n^{2}-21n+\frac{441}{4}=\frac{81}{4}$$

Factor n^{2}-21n+\frac{441}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n-\frac{21}{2}\right)^{2}=\frac{81}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{21}{2}\right)^{2}}=\sqrt{\frac{81}{4}}$$

Simplify.

$$n-\frac{21}{2}=\frac{9}{2} n-\frac{21}{2}=-\frac{9}{2}$$

Add \frac{21}{2} to both sides of the equation.

$$n=15 n=6$$