Solve 1500+1500(1+x)+1500(1+x)(1+x)=2160

$$1500+1500(1+x)+1500(1+x)(1+x)=2160$$

Solve for x

$$x=\frac{\sqrt{69}}{10}-\frac{3}{2}\approx -0.669337614$$

$$x=-\frac{\sqrt{69}}{10}-\frac{3}{2}\approx -2.330662386$$

Steps Using the Quadratic Formula

Multiply 1+x and 1+x to get \left(1+x\right)^{2}.

$$1500+1500\left(1+x\right)+1500\left(1+x\right)^{2}=2160$$

Use the distributive property to multiply 1500 by 1+x.

$$1500+1500+1500x+1500\left(1+x\right)^{2}=2160$$

Add 1500 and 1500 to get 3000.

$$3000+1500x+1500\left(1+x\right)^{2}=2160$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+x\right)^{2}.

$$3000+1500x+1500\left(1+2x+x^{2}\right)=2160$$

Use the distributive property to multiply 1500 by 1+2x+x^{2}.

$$3000+1500x+1500+3000x+1500x^{2}=2160$$

Add 3000 and 1500 to get 4500.

$$4500+1500x+3000x+1500x^{2}=2160$$

Combine 1500x and 3000x to get 4500x.

$$4500+4500x+1500x^{2}=2160$$

Subtract 2160 from both sides.

$$4500+4500x+1500x^{2}-2160=0$$

Subtract 2160 from 4500 to get 2340.

$$2340+4500x+1500x^{2}=0$$

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.

$$1500x^{2}+4500x+2340=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1500 for a, 4500 for b, and 2340 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4500±\sqrt{4500^{2}-4\times 1500\times 2340}}{2\times 1500}$$

Square 4500.

$$x=\frac{-4500±\sqrt{20250000-4\times 1500\times 2340}}{2\times 1500}$$

Multiply -4 times 1500.

$$x=\frac{-4500±\sqrt{20250000-6000\times 2340}}{2\times 1500}$$

Multiply -6000 times 2340.

$$x=\frac{-4500±\sqrt{20250000-14040000}}{2\times 1500}$$

Add 20250000 to -14040000.

$$x=\frac{-4500±\sqrt{6210000}}{2\times 1500}$$

Take the square root of 6210000.

$$x=\frac{-4500±300\sqrt{69}}{2\times 1500}$$

Multiply 2 times 1500.

$$x=\frac{-4500±300\sqrt{69}}{3000}$$

Now solve the equation x=\frac{-4500±300\sqrt{69}}{3000} when ± is plus. Add -4500 to 300\sqrt{69}.

$$x=\frac{300\sqrt{69}-4500}{3000}$$

Divide -4500+300\sqrt{69} by 3000.

$$x=\frac{\sqrt{69}}{10}-\frac{3}{2}$$

Now solve the equation x=\frac{-4500±300\sqrt{69}}{3000} when ± is minus. Subtract 300\sqrt{69} from -4500.

$$x=\frac{-300\sqrt{69}-4500}{3000}$$

Divide -4500-300\sqrt{69} by 3000.

$$x=-\frac{\sqrt{69}}{10}-\frac{3}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{69}}{10}-\frac{3}{2} x=-\frac{\sqrt{69}}{10}-\frac{3}{2}$$

Steps for Completing the Square

Multiply 1+x and 1+x to get \left(1+x\right)^{2}.

$$1500+1500\left(1+x\right)+1500\left(1+x\right)^{2}=2160$$

Use the distributive property to multiply 1500 by 1+x.

$$1500+1500+1500x+1500\left(1+x\right)^{2}=2160$$

Add 1500 and 1500 to get 3000.

$$3000+1500x+1500\left(1+x\right)^{2}=2160$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+x\right)^{2}.

$$3000+1500x+1500\left(1+2x+x^{2}\right)=2160$$

Use the distributive property to multiply 1500 by 1+2x+x^{2}.

$$3000+1500x+1500+3000x+1500x^{2}=2160$$

Add 3000 and 1500 to get 4500.

$$4500+1500x+3000x+1500x^{2}=2160$$

Combine 1500x and 3000x to get 4500x.

$$4500+4500x+1500x^{2}=2160$$

Subtract 4500 from both sides.

$$4500x+1500x^{2}=2160-4500$$

Subtract 4500 from 2160 to get -2340.

$$4500x+1500x^{2}=-2340$$

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.

$$1500x^{2}+4500x=-2340$$

Divide both sides by 1500.

$$\frac{1500x^{2}+4500x}{1500}=-\frac{2340}{1500}$$

Dividing by 1500 undoes the multiplication by 1500.

$$x^{2}+\frac{4500}{1500}x=-\frac{2340}{1500}$$

Divide 4500 by 1500.

$$x^{2}+3x=-\frac{2340}{1500}$$

Reduce the fraction \frac{-2340}{1500} to lowest terms by extracting and canceling out 60.

$$x^{2}+3x=-\frac{39}{25}$$

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.

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=-\frac{39}{25}+\left(\frac{3}{2}\right)^{2}$$

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

$$x^{2}+3x+\frac{9}{4}=-\frac{39}{25}+\frac{9}{4}$$

Add -\frac{39}{25} to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+3x+\frac{9}{4}=\frac{69}{100}$$

Factor x^{2}+3x+\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}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{69}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{69}{100}}$$

Simplify.

$$x+\frac{3}{2}=\frac{\sqrt{69}}{10} x+\frac{3}{2}=-\frac{\sqrt{69}}{10}$$

Subtract \frac{3}{2} from both sides of the equation.

$$x=\frac{\sqrt{69}}{10}-\frac{3}{2} x=-\frac{\sqrt{69}}{10}-\frac{3}{2}$$