Solve 15000=10000left(1+xright)^2

$$15000 = 10000 { \left(1+x \right) }^{ 2 }$$

Solve for x

$$x=\frac{\sqrt{6}}{2}-1\approx 0.224744871$$

$$x=-\frac{\sqrt{6}}{2}-1\approx -2.224744871$$

Steps Using the Quadratic Formula

Divide both sides by 10000.

$$\frac{15000}{10000}=\left(1+x\right)^{2}$$

Reduce the fraction \frac{15000}{10000} to lowest terms by extracting and canceling out 5000.

$$\frac{3}{2}=\left(1+x\right)^{2}$$

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

$$\frac{3}{2}=1+2x+x^{2}$$

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

$$1+2x+x^{2}=\frac{3}{2}$$

Subtract \frac{3}{2} from both sides.

$$1+2x+x^{2}-\frac{3}{2}=0$$

Subtract \frac{3}{2} from 1 to get -\frac{1}{2}.

$$-\frac{1}{2}+2x+x^{2}=0$$

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.

$$x^{2}+2x-\frac{1}{2}=0$$

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

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

Square 2.

$$x=\frac{-2±\sqrt{4-4\left(-\frac{1}{2}\right)}}{2}$$

Multiply -4 times -\frac{1}{2}.

$$x=\frac{-2±\sqrt{4+2}}{2}$$

Add 4 to 2.

$$x=\frac{-2±\sqrt{6}}{2}$$

Now solve the equation x=\frac{-2±\sqrt{6}}{2} when ± is plus. Add -2 to \sqrt{6}.

$$x=\frac{\sqrt{6}-2}{2}$$

Divide -2+\sqrt{6} by 2.

$$x=\frac{\sqrt{6}}{2}-1$$

Now solve the equation x=\frac{-2±\sqrt{6}}{2} when ± is minus. Subtract \sqrt{6} from -2.

$$x=\frac{-\sqrt{6}-2}{2}$$

Divide -2-\sqrt{6} by 2.

$$x=-\frac{\sqrt{6}}{2}-1$$

The equation is now solved.

$$x=\frac{\sqrt{6}}{2}-1 x=-\frac{\sqrt{6}}{2}-1$$

Steps for Completing the Square

Divide both sides by 10000.

$$\frac{15000}{10000}=\left(1+x\right)^{2}$$

Reduce the fraction \frac{15000}{10000} to lowest terms by extracting and canceling out 5000.

$$\frac{3}{2}=\left(1+x\right)^{2}$$

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

$$\frac{3}{2}=1+2x+x^{2}$$

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

$$1+2x+x^{2}=\frac{3}{2}$$

Subtract 1 from both sides.

$$2x+x^{2}=\frac{3}{2}-1$$

Subtract 1 from \frac{3}{2} to get \frac{1}{2}.

$$2x+x^{2}=\frac{1}{2}$$

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.

$$x^{2}+2x=\frac{1}{2}$$

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+2x+1^{2}=\frac{1}{2}+1^{2}$$

Square 1.

$$x^{2}+2x+1=\frac{1}{2}+1$$

Add \frac{1}{2} to 1.

$$x^{2}+2x+1=\frac{3}{2}$$

Factor x^{2}+2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+1\right)^{2}=\frac{3}{2}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{\frac{3}{2}}$$

Simplify.

$$x+1=\frac{\sqrt{6}}{2} x+1=-\frac{\sqrt{6}}{2}$$

Subtract 1 from both sides of the equation.

$$x=\frac{\sqrt{6}}{2}-1 x=-\frac{\sqrt{6}}{2}-1$$