Solve 100(0.8+x)(0.8+x)=135.2

$$100(0.8+x)(0.8+x)=135.2$$

Solve for x

$$x=\frac{13\sqrt{5}}{25}-0.8\approx 0.362755348$$

$$x=-\frac{13\sqrt{5}}{25}-0.8\approx -1.962755348$$

Steps Using the Quadratic Formula

Multiply 0.8+x and 0.8+x to get \left(0.8+x\right)^{2}.

$$100\left(0.8+x\right)^{2}=135.2$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(0.8+x\right)^{2}.

$$100\left(0.64+1.6x+x^{2}\right)=135.2$$

Use the distributive property to multiply 100 by 0.64+1.6x+x^{2}.

$$64+160x+100x^{2}=135.2$$

Subtract 135.2 from both sides.

$$64+160x+100x^{2}-135.2=0$$

Subtract 135.2 from 64 to get -71.2.

$$-71.2+160x+100x^{2}=0$$

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.

$$100x^{2}+160x-71.2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 100 for a, 160 for b, and -71.2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-160±\sqrt{160^{2}-4\times 100\left(-71.2\right)}}{2\times 100}$$

Square 160.

$$x=\frac{-160±\sqrt{25600-4\times 100\left(-71.2\right)}}{2\times 100}$$

Multiply -4 times 100.

$$x=\frac{-160±\sqrt{25600-400\left(-71.2\right)}}{2\times 100}$$

Multiply -400 times -71.2.

$$x=\frac{-160±\sqrt{25600+28480}}{2\times 100}$$

Add 25600 to 28480.

$$x=\frac{-160±\sqrt{54080}}{2\times 100}$$

Take the square root of 54080.

$$x=\frac{-160±104\sqrt{5}}{2\times 100}$$

Multiply 2 times 100.

$$x=\frac{-160±104\sqrt{5}}{200}$$

Now solve the equation x=\frac{-160±104\sqrt{5}}{200} when ± is plus. Add -160 to 104\sqrt{5}.

$$x=\frac{104\sqrt{5}-160}{200}$$

Divide -160+104\sqrt{5} by 200.

$$x=\frac{13\sqrt{5}}{25}-\frac{4}{5}$$

Now solve the equation x=\frac{-160±104\sqrt{5}}{200} when ± is minus. Subtract 104\sqrt{5} from -160.

$$x=\frac{-104\sqrt{5}-160}{200}$$

Divide -160-104\sqrt{5} by 200.

$$x=-\frac{13\sqrt{5}}{25}-\frac{4}{5}$$

The equation is now solved.

$$x=\frac{13\sqrt{5}}{25}-\frac{4}{5} x=-\frac{13\sqrt{5}}{25}-\frac{4}{5}$$

Steps for Completing the Square

Multiply 0.8+x and 0.8+x to get \left(0.8+x\right)^{2}.

$$100\left(0.8+x\right)^{2}=135.2$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(0.8+x\right)^{2}.

$$100\left(0.64+1.6x+x^{2}\right)=135.2$$

Use the distributive property to multiply 100 by 0.64+1.6x+x^{2}.

$$64+160x+100x^{2}=135.2$$

Subtract 64 from both sides.

$$160x+100x^{2}=135.2-64$$

Subtract 64 from 135.2 to get 71.2.

$$160x+100x^{2}=71.2$$

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.

$$100x^{2}+160x=71.2$$

Divide both sides by 100.

$$\frac{100x^{2}+160x}{100}=\frac{71.2}{100}$$

Dividing by 100 undoes the multiplication by 100.

$$x^{2}+\frac{160}{100}x=\frac{71.2}{100}$$

Reduce the fraction \frac{160}{100} to lowest terms by extracting and canceling out 20.

$$x^{2}+\frac{8}{5}x=\frac{71.2}{100}$$

Divide 71.2 by 100.

$$x^{2}+\frac{8}{5}x=0.712$$

Divide \frac{8}{5}, the coefficient of the x term, by 2 to get \frac{4}{5}. Then add the square of \frac{4}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{8}{5}x+\left(\frac{4}{5}\right)^{2}=0.712+\left(\frac{4}{5}\right)^{2}$$

Square \frac{4}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{8}{5}x+\frac{16}{25}=0.712+\frac{16}{25}$$

Add 0.712 to \frac{16}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{8}{5}x+\frac{16}{25}=\frac{169}{125}$$

Factor x^{2}+\frac{8}{5}x+\frac{16}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{4}{5}\right)^{2}=\frac{169}{125}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{4}{5}\right)^{2}}=\sqrt{\frac{169}{125}}$$

Simplify.

$$x+\frac{4}{5}=\frac{13\sqrt{5}}{25} x+\frac{4}{5}=-\frac{13\sqrt{5}}{25}$$

Subtract \frac{4}{5} from both sides of the equation.

$$x=\frac{13\sqrt{5}}{25}-\frac{4}{5} x=-\frac{13\sqrt{5}}{25}-\frac{4}{5}$$