Solve 100(60-35)(1-x)^2=3600

$$100 ( 60 - 35 ) ( 1 - x ) ^ { 2 } = 3600$$

Solve for x

$$x=-\frac{1}{5}=-0.2$$

$$x = \frac{11}{5} = 2\frac{1}{5} = 2.2$$

Steps Using Factoring By Grouping

Subtract 35 from 60 to get 25.

$$100\times 25\left(1-x\right)^{2}=3600$$

Multiply 100 and 25 to get 2500.

$$2500\left(1-x\right)^{2}=3600$$

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

$$2500\left(1-2x+x^{2}\right)=3600$$

Use the distributive property to multiply 2500 by 1-2x+x^{2}.

$$2500-5000x+2500x^{2}=3600$$

Subtract 3600 from both sides.

$$2500-5000x+2500x^{2}-3600=0$$

Subtract 3600 from 2500 to get -1100.

$$-1100-5000x+2500x^{2}=0$$

Divide both sides by 100.

$$-11-50x+25x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$25x^{2}-50x-11=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 25x^{2}+ax+bx-11. To find a and b, set up a system to be solved.

$$a+b=-50 ab=25\left(-11\right)=-275$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -275.

$$1,-275 5,-55 11,-25$$

Calculate the sum for each pair.

$$1-275=-274 5-55=-50 11-25=-14$$

The solution is the pair that gives sum -50.

$$a=-55 b=5$$

Rewrite 25x^{2}-50x-11 as \left(25x^{2}-55x\right)+\left(5x-11\right).

$$\left(25x^{2}-55x\right)+\left(5x-11\right)$$

Factor out 5x in 25x^{2}-55x.

$$5x\left(5x-11\right)+5x-11$$

Factor out common term 5x-11 by using distributive property.

$$\left(5x-11\right)\left(5x+1\right)$$

To find equation solutions, solve 5x-11=0 and 5x+1=0.

$$x=\frac{11}{5} x=-\frac{1}{5}$$

Steps Using the Quadratic Formula

Subtract 35 from 60 to get 25.

$$100\times 25\left(1-x\right)^{2}=3600$$

Multiply 100 and 25 to get 2500.

$$2500\left(1-x\right)^{2}=3600$$

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

$$2500\left(1-2x+x^{2}\right)=3600$$

Use the distributive property to multiply 2500 by 1-2x+x^{2}.

$$2500-5000x+2500x^{2}=3600$$

Subtract 3600 from both sides.

$$2500-5000x+2500x^{2}-3600=0$$

Subtract 3600 from 2500 to get -1100.

$$-1100-5000x+2500x^{2}=0$$

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.

$$2500x^{2}-5000x-1100=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2500 for a, -5000 for b, and -1100 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5000\right)±\sqrt{\left(-5000\right)^{2}-4\times 2500\left(-1100\right)}}{2\times 2500}$$

Square -5000.

$$x=\frac{-\left(-5000\right)±\sqrt{25000000-4\times 2500\left(-1100\right)}}{2\times 2500}$$

Multiply -4 times 2500.

$$x=\frac{-\left(-5000\right)±\sqrt{25000000-10000\left(-1100\right)}}{2\times 2500}$$

Multiply -10000 times -1100.

$$x=\frac{-\left(-5000\right)±\sqrt{25000000+11000000}}{2\times 2500}$$

Add 25000000 to 11000000.

$$x=\frac{-\left(-5000\right)±\sqrt{36000000}}{2\times 2500}$$

Take the square root of 36000000.

$$x=\frac{-\left(-5000\right)±6000}{2\times 2500}$$

The opposite of -5000 is 5000.

$$x=\frac{5000±6000}{2\times 2500}$$

Multiply 2 times 2500.

$$x=\frac{5000±6000}{5000}$$

Now solve the equation x=\frac{5000±6000}{5000} when ± is plus. Add 5000 to 6000.

$$x=\frac{11000}{5000}$$

Reduce the fraction \frac{11000}{5000} to lowest terms by extracting and canceling out 1000.

$$x=\frac{11}{5}$$

Now solve the equation x=\frac{5000±6000}{5000} when ± is minus. Subtract 6000 from 5000.

$$x=-\frac{1000}{5000}$$

Reduce the fraction \frac{-1000}{5000} to lowest terms by extracting and canceling out 1000.

$$x=-\frac{1}{5}$$

The equation is now solved.

$$x=\frac{11}{5} x=-\frac{1}{5}$$

Steps for Completing the Square

Subtract 35 from 60 to get 25.

$$100\times 25\left(1-x\right)^{2}=3600$$

Multiply 100 and 25 to get 2500.

$$2500\left(1-x\right)^{2}=3600$$

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

$$2500\left(1-2x+x^{2}\right)=3600$$

Use the distributive property to multiply 2500 by 1-2x+x^{2}.

$$2500-5000x+2500x^{2}=3600$$

Subtract 2500 from both sides.

$$-5000x+2500x^{2}=3600-2500$$

Subtract 2500 from 3600 to get 1100.

$$-5000x+2500x^{2}=1100$$

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.

$$2500x^{2}-5000x=1100$$

Divide both sides by 2500.

$$\frac{2500x^{2}-5000x}{2500}=\frac{1100}{2500}$$

Dividing by 2500 undoes the multiplication by 2500.

$$x^{2}+\left(-\frac{5000}{2500}\right)x=\frac{1100}{2500}$$

Divide -5000 by 2500.

$$x^{2}-2x=\frac{1100}{2500}$$

Reduce the fraction \frac{1100}{2500} to lowest terms by extracting and canceling out 100.

$$x^{2}-2x=\frac{11}{25}$$

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.

$$x^{2}-2x+1=\frac{11}{25}+1$$

Add \frac{11}{25} to 1.

$$x^{2}-2x+1=\frac{36}{25}$$

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}.

$$\left(x-1\right)^{2}=\frac{36}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{\frac{36}{25}}$$

Simplify.

$$x-1=\frac{6}{5} x-1=-\frac{6}{5}$$

Add 1 to both sides of the equation.

$$x=\frac{11}{5} x=-\frac{1}{5}$$