Solve 44=n(n-3)/2

$$44 = \frac { n ( n - 3 ) } { 2 }$$

Solve for n

$$n=-8$$

$$n=11$$

Steps Using the Quadratic Formula

Multiply both sides by 2.

$$44\times 2=n\left(n-3\right)$$

Multiply 44 and 2 to get 88.

$$88=n\left(n-3\right)$$

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

$$88=n^{2}-3n$$

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

$$n^{2}-3n=88$$

Subtract 88 from both sides.

$$n^{2}-3n-88=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -3 for b, and -88 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-88\right)}}{2}$$

Square -3.

$$n=\frac{-\left(-3\right)±\sqrt{9-4\left(-88\right)}}{2}$$

Multiply -4 times -88.

$$n=\frac{-\left(-3\right)±\sqrt{9+352}}{2}$$

Add 9 to 352.

$$n=\frac{-\left(-3\right)±\sqrt{361}}{2}$$

Take the square root of 361.

$$n=\frac{-\left(-3\right)±19}{2}$$

The opposite of -3 is 3.

$$n=\frac{3±19}{2}$$

Now solve the equation n=\frac{3±19}{2} when ± is plus. Add 3 to 19.

$$n=\frac{22}{2}$$

Divide 22 by 2.

$$n=11$$

Now solve the equation n=\frac{3±19}{2} when ± is minus. Subtract 19 from 3.

$$n=-\frac{16}{2}$$

Divide -16 by 2.

$$n=-8$$

The equation is now solved.

$$n=11 n=-8$$

Steps for Completing the Square

Multiply both sides by 2.

$$44\times 2=n\left(n-3\right)$$

Multiply 44 and 2 to get 88.

$$88=n\left(n-3\right)$$

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

$$88=n^{2}-3n$$

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

$$n^{2}-3n=88$$

Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}-3n+\left(-\frac{3}{2}\right)^{2}=88+\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$n^{2}-3n+\frac{9}{4}=88+\frac{9}{4}$$

Add 88 to \frac{9}{4}.

$$n^{2}-3n+\frac{9}{4}=\frac{361}{4}$$

Factor n^{2}-3n+\frac{9}{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}.

$$\left(n-\frac{3}{2}\right)^{2}=\frac{361}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{3}{2}\right)^{2}}=\sqrt{\frac{361}{4}}$$

Simplify.

$$n-\frac{3}{2}=\frac{19}{2} n-\frac{3}{2}=-\frac{19}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$n=11 n=-8$$