Solve 606=15000{left(1+y/2right)}^2

$$606 = 15000 { \left(1+ \frac{ y }{ 2 } \right) }^{ 2 }$$

Solve for y

$$y=\frac{\sqrt{101}}{25}-2\approx -1.598004975$$

$$y=-\frac{\sqrt{101}}{25}-2\approx -2.401995025$$

Steps Using the Quadratic Formula

Divide both sides by 15000.

$$\frac{606}{15000}=\left(1+\frac{y}{2}\right)^{2}$$

Reduce the fraction \frac{606}{15000} to lowest terms by extracting and canceling out 6.

$$\frac{101}{2500}=\left(1+\frac{y}{2}\right)^{2}$$

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

$$\frac{101}{2500}=1+2\times \frac{y}{2}+\left(\frac{y}{2}\right)^{2}$$

Express 2\times \frac{y}{2} as a single fraction.

$$\frac{101}{2500}=1+\frac{2y}{2}+\left(\frac{y}{2}\right)^{2}$$

Cancel out 2 and 2.

$$\frac{101}{2500}=1+y+\left(\frac{y}{2}\right)^{2}$$

To raise \frac{y}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{101}{2500}=1+y+\frac{y^{2}}{2^{2}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1+y times \frac{2^{2}}{2^{2}}.

$$\frac{101}{2500}=\frac{\left(1+y\right)\times 2^{2}}{2^{2}}+\frac{y^{2}}{2^{2}}$$

Since \frac{\left(1+y\right)\times 2^{2}}{2^{2}} and \frac{y^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$\frac{101}{2500}=\frac{\left(1+y\right)\times 2^{2}+y^{2}}{2^{2}}$$

Do the multiplications in \left(1+y\right)\times 2^{2}+y^{2}.

$$\frac{101}{2500}=\frac{4+4y+y^{2}}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{101}{2500}=\frac{4+4y+y^{2}}{4}$$

Divide each term of 4+4y+y^{2} by 4 to get 1+y+\frac{1}{4}y^{2}.

$$\frac{101}{2500}=1+y+\frac{1}{4}y^{2}$$

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

$$1+y+\frac{1}{4}y^{2}=\frac{101}{2500}$$

Subtract \frac{101}{2500} from both sides.

$$1+y+\frac{1}{4}y^{2}-\frac{101}{2500}=0$$

Subtract \frac{101}{2500} from 1 to get \frac{2399}{2500}.

$$\frac{2399}{2500}+y+\frac{1}{4}y^{2}=0$$

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.

$$\frac{1}{4}y^{2}+y+\frac{2399}{2500}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{4} for a, 1 for b, and \frac{2399}{2500} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-1±\sqrt{1^{2}-4\times \frac{1}{4}\times \frac{2399}{2500}}}{2\times \frac{1}{4}}$$

Square 1.

$$y=\frac{-1±\sqrt{1-4\times \frac{1}{4}\times \frac{2399}{2500}}}{2\times \frac{1}{4}}$$

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

$$y=\frac{-1±\sqrt{1-\frac{2399}{2500}}}{2\times \frac{1}{4}}$$

Add 1 to -\frac{2399}{2500}.

$$y=\frac{-1±\sqrt{\frac{101}{2500}}}{2\times \frac{1}{4}}$$

Take the square root of \frac{101}{2500}.

$$y=\frac{-1±\frac{\sqrt{101}}{50}}{2\times \frac{1}{4}}$$

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

$$y=\frac{-1±\frac{\sqrt{101}}{50}}{\frac{1}{2}}$$

Now solve the equation y=\frac{-1±\frac{\sqrt{101}}{50}}{\frac{1}{2}} when ± is plus. Add -1 to \frac{\sqrt{101}}{50}.

$$y=\frac{\frac{\sqrt{101}}{50}-1}{\frac{1}{2}}$$

Divide -1+\frac{\sqrt{101}}{50} by \frac{1}{2} by multiplying -1+\frac{\sqrt{101}}{50} by the reciprocal of \frac{1}{2}.

$$y=\frac{\sqrt{101}}{25}-2$$

Now solve the equation y=\frac{-1±\frac{\sqrt{101}}{50}}{\frac{1}{2}} when ± is minus. Subtract \frac{\sqrt{101}}{50} from -1.

$$y=\frac{-\frac{\sqrt{101}}{50}-1}{\frac{1}{2}}$$

Divide -1-\frac{\sqrt{101}}{50} by \frac{1}{2} by multiplying -1-\frac{\sqrt{101}}{50} by the reciprocal of \frac{1}{2}.

$$y=-\frac{\sqrt{101}}{25}-2$$

The equation is now solved.

$$y=\frac{\sqrt{101}}{25}-2 y=-\frac{\sqrt{101}}{25}-2$$

Steps for Completing the Square

Divide both sides by 15000.

$$\frac{606}{15000}=\left(1+\frac{y}{2}\right)^{2}$$

Reduce the fraction \frac{606}{15000} to lowest terms by extracting and canceling out 6.

$$\frac{101}{2500}=\left(1+\frac{y}{2}\right)^{2}$$

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

$$\frac{101}{2500}=1+2\times \frac{y}{2}+\left(\frac{y}{2}\right)^{2}$$

Express 2\times \frac{y}{2} as a single fraction.

$$\frac{101}{2500}=1+\frac{2y}{2}+\left(\frac{y}{2}\right)^{2}$$

Cancel out 2 and 2.

$$\frac{101}{2500}=1+y+\left(\frac{y}{2}\right)^{2}$$

To raise \frac{y}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{101}{2500}=1+y+\frac{y^{2}}{2^{2}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1+y times \frac{2^{2}}{2^{2}}.

$$\frac{101}{2500}=\frac{\left(1+y\right)\times 2^{2}}{2^{2}}+\frac{y^{2}}{2^{2}}$$

Since \frac{\left(1+y\right)\times 2^{2}}{2^{2}} and \frac{y^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$\frac{101}{2500}=\frac{\left(1+y\right)\times 2^{2}+y^{2}}{2^{2}}$$

Do the multiplications in \left(1+y\right)\times 2^{2}+y^{2}.

$$\frac{101}{2500}=\frac{4+4y+y^{2}}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{101}{2500}=\frac{4+4y+y^{2}}{4}$$

Divide each term of 4+4y+y^{2} by 4 to get 1+y+\frac{1}{4}y^{2}.

$$\frac{101}{2500}=1+y+\frac{1}{4}y^{2}$$

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

$$1+y+\frac{1}{4}y^{2}=\frac{101}{2500}$$

Subtract 1 from both sides.

$$y+\frac{1}{4}y^{2}=\frac{101}{2500}-1$$

Subtract 1 from \frac{101}{2500} to get -\frac{2399}{2500}.

$$y+\frac{1}{4}y^{2}=-\frac{2399}{2500}$$

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.

$$\frac{1}{4}y^{2}+y=-\frac{2399}{2500}$$

Multiply both sides by 4.

$$\frac{\frac{1}{4}y^{2}+y}{\frac{1}{4}}=-\frac{\frac{2399}{2500}}{\frac{1}{4}}$$

Dividing by \frac{1}{4} undoes the multiplication by \frac{1}{4}.

$$y^{2}+\frac{1}{\frac{1}{4}}y=-\frac{\frac{2399}{2500}}{\frac{1}{4}}$$

Divide 1 by \frac{1}{4} by multiplying 1 by the reciprocal of \frac{1}{4}.

$$y^{2}+4y=-\frac{\frac{2399}{2500}}{\frac{1}{4}}$$

Divide -\frac{2399}{2500} by \frac{1}{4} by multiplying -\frac{2399}{2500} by the reciprocal of \frac{1}{4}.

$$y^{2}+4y=-\frac{2399}{625}$$

Divide 4, the coefficient of the x term, by 2 to get 2. Then add the square of 2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}+4y+2^{2}=-\frac{2399}{625}+2^{2}$$

Square 2.

$$y^{2}+4y+4=-\frac{2399}{625}+4$$

Add -\frac{2399}{625} to 4.

$$y^{2}+4y+4=\frac{101}{625}$$

Factor y^{2}+4y+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}.

$$\left(y+2\right)^{2}=\frac{101}{625}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+2\right)^{2}}=\sqrt{\frac{101}{625}}$$

Simplify.

$$y+2=\frac{\sqrt{101}}{25} y+2=-\frac{\sqrt{101}}{25}$$

Subtract 2 from both sides of the equation.

$$y=\frac{\sqrt{101}}{25}-2 y=-\frac{\sqrt{101}}{25}-2$$