Solve 100(1-20%)(1+x)^2=135.2

$$100 ( 1 - 20 \% ) ( 1 + x ) ^ { 2 } = 135.2$$

Solve for x

$$x=0.3$$

$$x=-2.3$$

Steps Using the Quadratic Formula

Reduce the fraction \frac{20}{100} to lowest terms by extracting and canceling out 20.

$$100\left(1-\frac{1}{5}\right)\left(1+x\right)^{2}=135.2$$

Subtract \frac{1}{5} from 1 to get \frac{4}{5}.

$$100\times \frac{4}{5}\left(1+x\right)^{2}=135.2$$

Multiply 100 and \frac{4}{5} to get 80.

$$80\left(1+x\right)^{2}=135.2$$

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

$$80\left(1+2x+x^{2}\right)=135.2$$

Use the distributive property to multiply 80 by 1+2x+x^{2}.

$$80+160x+80x^{2}=135.2$$

Subtract 135.2 from both sides.

$$80+160x+80x^{2}-135.2=0$$

Subtract 135.2 from 80 to get -55.2.

$$-55.2+160x+80x^{2}=0$$

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.

$$80x^{2}+160x-55.2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 80 for a, 160 for b, and -55.2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-160±\sqrt{160^{2}-4\times 80\left(-55.2\right)}}{2\times 80}$$

Square 160.

$$x=\frac{-160±\sqrt{25600-4\times 80\left(-55.2\right)}}{2\times 80}$$

Multiply -4 times 80.

$$x=\frac{-160±\sqrt{25600-320\left(-55.2\right)}}{2\times 80}$$

Multiply -320 times -55.2.

$$x=\frac{-160±\sqrt{25600+17664}}{2\times 80}$$

Add 25600 to 17664.

$$x=\frac{-160±\sqrt{43264}}{2\times 80}$$

Take the square root of 43264.

$$x=\frac{-160±208}{2\times 80}$$

Multiply 2 times 80.

$$x=\frac{-160±208}{160}$$

Now solve the equation x=\frac{-160±208}{160} when ± is plus. Add -160 to 208.

$$x=\frac{48}{160}$$

Reduce the fraction \frac{48}{160} to lowest terms by extracting and canceling out 16.

$$x=\frac{3}{10}$$

Now solve the equation x=\frac{-160±208}{160} when ± is minus. Subtract 208 from -160.

$$x=-\frac{368}{160}$$

Reduce the fraction \frac{-368}{160} to lowest terms by extracting and canceling out 16.

$$x=-\frac{23}{10}$$

The equation is now solved.

$$x=\frac{3}{10} x=-\frac{23}{10}$$

Steps for Completing the Square

Reduce the fraction \frac{20}{100} to lowest terms by extracting and canceling out 20.

$$100\left(1-\frac{1}{5}\right)\left(1+x\right)^{2}=135.2$$

Subtract \frac{1}{5} from 1 to get \frac{4}{5}.

$$100\times \frac{4}{5}\left(1+x\right)^{2}=135.2$$

Multiply 100 and \frac{4}{5} to get 80.

$$80\left(1+x\right)^{2}=135.2$$

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

$$80\left(1+2x+x^{2}\right)=135.2$$

Use the distributive property to multiply 80 by 1+2x+x^{2}.

$$80+160x+80x^{2}=135.2$$

Subtract 80 from both sides.

$$160x+80x^{2}=135.2-80$$

Subtract 80 from 135.2 to get 55.2.

$$160x+80x^{2}=55.2$$

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.

$$80x^{2}+160x=55.2$$

Divide both sides by 80.

$$\frac{80x^{2}+160x}{80}=\frac{55.2}{80}$$

Dividing by 80 undoes the multiplication by 80.

$$x^{2}+\frac{160}{80}x=\frac{55.2}{80}$$

Divide 160 by 80.

$$x^{2}+2x=\frac{55.2}{80}$$

Divide 55.2 by 80.

$$x^{2}+2x=0.69$$

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.

$$x^{2}+2x+1^{2}=0.69+1^{2}$$

Square 1.

$$x^{2}+2x+1=0.69+1$$

Add 0.69 to 1.

$$x^{2}+2x+1=1.69$$

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}.

$$\left(x+1\right)^{2}=1.69$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{1.69}$$

Simplify.

$$x+1=\frac{13}{10} x+1=-\frac{13}{10}$$

Subtract 1 from both sides of the equation.

$$x=\frac{3}{10} x=-\frac{23}{10}$$