Solve 54=[2(-12)+(n-1)3]n/2

$$54 = \frac { [ 2 ( - 12 ) + ( n - 1 ) 3 ] n } { 2 }$$

Solve for n

$$n=-3$$

$$n=12$$

Steps Using the Quadratic Formula

Multiply both sides by 2.

$$54\times 2=\left(2\left(-12\right)+\left(n-1\right)\times 3\right)n$$

Multiply 54 and 2 to get 108.

$$108=\left(2\left(-12\right)+\left(n-1\right)\times 3\right)n$$

Multiply 2 and -12 to get -24.

$$108=\left(-24+\left(n-1\right)\times 3\right)n$$

Use the distributive property to multiply n-1 by 3.

$$108=\left(-24+3n-3\right)n$$

Subtract 3 from -24 to get -27.

$$108=\left(-27+3n\right)n$$

Use the distributive property to multiply -27+3n by n.

$$108=-27n+3n^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$-27n+3n^{2}=108$$

Subtract 108 from both sides.

$$-27n+3n^{2}-108=0$$

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.

$$3n^{2}-27n-108=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -27 for b, and -108 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-\left(-27\right)±\sqrt{\left(-27\right)^{2}-4\times 3\left(-108\right)}}{2\times 3}$$

Square -27.

$$n=\frac{-\left(-27\right)±\sqrt{729-4\times 3\left(-108\right)}}{2\times 3}$$

Multiply -4 times 3.

$$n=\frac{-\left(-27\right)±\sqrt{729-12\left(-108\right)}}{2\times 3}$$

Multiply -12 times -108.

$$n=\frac{-\left(-27\right)±\sqrt{729+1296}}{2\times 3}$$

Add 729 to 1296.

$$n=\frac{-\left(-27\right)±\sqrt{2025}}{2\times 3}$$

Take the square root of 2025.

$$n=\frac{-\left(-27\right)±45}{2\times 3}$$

The opposite of -27 is 27.

$$n=\frac{27±45}{2\times 3}$$

Multiply 2 times 3.

$$n=\frac{27±45}{6}$$

Now solve the equation n=\frac{27±45}{6} when ± is plus. Add 27 to 45.

$$n=\frac{72}{6}$$

Divide 72 by 6.

$$n=12$$

Now solve the equation n=\frac{27±45}{6} when ± is minus. Subtract 45 from 27.

$$n=-\frac{18}{6}$$

Divide -18 by 6.

$$n=-3$$

The equation is now solved.

$$n=12 n=-3$$

Steps for Completing the Square

Multiply both sides by 2.

$$54\times 2=\left(2\left(-12\right)+\left(n-1\right)\times 3\right)n$$

Multiply 54 and 2 to get 108.

$$108=\left(2\left(-12\right)+\left(n-1\right)\times 3\right)n$$

Multiply 2 and -12 to get -24.

$$108=\left(-24+\left(n-1\right)\times 3\right)n$$

Use the distributive property to multiply n-1 by 3.

$$108=\left(-24+3n-3\right)n$$

Subtract 3 from -24 to get -27.

$$108=\left(-27+3n\right)n$$

Use the distributive property to multiply -27+3n by n.

$$108=-27n+3n^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$-27n+3n^{2}=108$$

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.

$$3n^{2}-27n=108$$

Divide both sides by 3.

$$\frac{3n^{2}-27n}{3}=\frac{108}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$n^{2}+\left(-\frac{27}{3}\right)n=\frac{108}{3}$$

Divide -27 by 3.

$$n^{2}-9n=\frac{108}{3}$$

Divide 108 by 3.

$$n^{2}-9n=36$$

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

$$n^{2}-9n+\left(-\frac{9}{2}\right)^{2}=36+\left(-\frac{9}{2}\right)^{2}$$

Square -\frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$n^{2}-9n+\frac{81}{4}=36+\frac{81}{4}$$

Add 36 to \frac{81}{4}.

$$n^{2}-9n+\frac{81}{4}=\frac{225}{4}$$

Factor n^{2}-9n+\frac{81}{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}.

$$\left(n-\frac{9}{2}\right)^{2}=\frac{225}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{9}{2}\right)^{2}}=\sqrt{\frac{225}{4}}$$

Simplify.

$$n-\frac{9}{2}=\frac{15}{2} n-\frac{9}{2}=-\frac{15}{2}$$

Add \frac{9}{2} to both sides of the equation.

$$n=12 n=-3$$