Solve (120-(n-1)3)n/2=600

$$\frac { ( 120 - ( n - 1 ) 3 ) n } { 2 } = 600$$

Solve for n

$$n=16$$

$$n=25$$

Steps Using the Quadratic Formula

Multiply both sides by 2.

$$\left(120-\left(n-1\right)\times 3\right)n=600\times 2$$

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

$$\left(120-\left(3n-3\right)\right)n=600\times 2$$

To find the opposite of 3n-3, find the opposite of each term.

$$\left(120-3n-\left(-3\right)\right)n=600\times 2$$

The opposite of -3 is 3.

$$\left(120-3n+3\right)n=600\times 2$$

Add 120 and 3 to get 123.

$$\left(123-3n\right)n=600\times 2$$

Use the distributive property to multiply 123-3n by n.

$$123n-3n^{2}=600\times 2$$

Multiply 600 and 2 to get 1200.

$$123n-3n^{2}=1200$$

Subtract 1200 from both sides.

$$123n-3n^{2}-1200=0$$

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.

$$-3n^{2}+123n-1200=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 123 for b, and -1200 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-123±\sqrt{123^{2}-4\left(-3\right)\left(-1200\right)}}{2\left(-3\right)}$$

Square 123.

$$n=\frac{-123±\sqrt{15129-4\left(-3\right)\left(-1200\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$n=\frac{-123±\sqrt{15129+12\left(-1200\right)}}{2\left(-3\right)}$$

Multiply 12 times -1200.

$$n=\frac{-123±\sqrt{15129-14400}}{2\left(-3\right)}$$

Add 15129 to -14400.

$$n=\frac{-123±\sqrt{729}}{2\left(-3\right)}$$

Take the square root of 729.

$$n=\frac{-123±27}{2\left(-3\right)}$$

Multiply 2 times -3.

$$n=\frac{-123±27}{-6}$$

Now solve the equation n=\frac{-123±27}{-6} when ± is plus. Add -123 to 27.

$$n=-\frac{96}{-6}$$

Divide -96 by -6.

$$n=16$$

Now solve the equation n=\frac{-123±27}{-6} when ± is minus. Subtract 27 from -123.

$$n=-\frac{150}{-6}$$

Divide -150 by -6.

$$n=25$$

The equation is now solved.

$$n=16 n=25$$

Steps for Completing the Square

Multiply both sides by 2.

$$\left(120-\left(n-1\right)\times 3\right)n=600\times 2$$

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

$$\left(120-\left(3n-3\right)\right)n=600\times 2$$

To find the opposite of 3n-3, find the opposite of each term.

$$\left(120-3n-\left(-3\right)\right)n=600\times 2$$

The opposite of -3 is 3.

$$\left(120-3n+3\right)n=600\times 2$$

Add 120 and 3 to get 123.

$$\left(123-3n\right)n=600\times 2$$

Use the distributive property to multiply 123-3n by n.

$$123n-3n^{2}=600\times 2$$

Multiply 600 and 2 to get 1200.

$$123n-3n^{2}=1200$$

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.

$$-3n^{2}+123n=1200$$

Divide both sides by -3.

$$\frac{-3n^{2}+123n}{-3}=\frac{1200}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$n^{2}+\frac{123}{-3}n=\frac{1200}{-3}$$

Divide 123 by -3.

$$n^{2}-41n=\frac{1200}{-3}$$

Divide 1200 by -3.

$$n^{2}-41n=-400$$

Divide -41, the coefficient of the x term, by 2 to get -\frac{41}{2}. Then add the square of -\frac{41}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}-41n+\left(-\frac{41}{2}\right)^{2}=-400+\left(-\frac{41}{2}\right)^{2}$$

Square -\frac{41}{2} by squaring both the numerator and the denominator of the fraction.

$$n^{2}-41n+\frac{1681}{4}=-400+\frac{1681}{4}$$

Add -400 to \frac{1681}{4}.

$$n^{2}-41n+\frac{1681}{4}=\frac{81}{4}$$

Factor n^{2}-41n+\frac{1681}{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}.

$$\left(n-\frac{41}{2}\right)^{2}=\frac{81}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{41}{2}\right)^{2}}=\sqrt{\frac{81}{4}}$$

Simplify.

$$n-\frac{41}{2}=\frac{9}{2} n-\frac{41}{2}=-\frac{9}{2}$$

Add \frac{41}{2} to both sides of the equation.

$$n=25 n=16$$