Solve {left(2.5+x/2right)}^2-{left(2.5-x/2right)}^2={left(4left(2.5+5/2right)/5right)}^2

$${ \left(2.5+ \frac{ x }{ 2 } \right) }^{ 2 } - { \left(2.5- \frac{ x }{ 2 } \right) }^{ 2 } = { \left( \frac{ 4 \left( 2.5+ \frac{ 5 }{ 2 } \right) }{ 5 } \right) }^{ 2 }$$

Solve for x

$$x=3.2$$

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2.5+\frac{x}{2}\right)^{2}.

$$6.25+5\times \frac{x}{2}+\left(\frac{x}{2}\right)^{2}-\left(2.5-\frac{x}{2}\right)^{2}=\left(\frac{4\left(2.5+\frac{5}{2}\right)}{5}\right)^{2}$$

Express 5\times \frac{x}{2} as a single fraction.

$$6.25+\frac{5x}{2}+\left(\frac{x}{2}\right)^{2}-\left(2.5-\frac{x}{2}\right)^{2}=\left(\frac{4\left(2.5+\frac{5}{2}\right)}{5}\right)^{2}$$

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

$$6.25+\frac{5x}{2}+\frac{x^{2}}{2^{2}}-\left(2.5-\frac{x}{2}\right)^{2}=\left(\frac{4\left(2.5+\frac{5}{2}\right)}{5}\right)^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2 and 2^{2} is 4. Multiply \frac{5x}{2} times \frac{2}{2}.

$$6.25+\frac{2\times 5x}{4}+\frac{x^{2}}{4}-\left(2.5-\frac{x}{2}\right)^{2}=\left(\frac{4\left(2.5+\frac{5}{2}\right)}{5}\right)^{2}$$

Since \frac{2\times 5x}{4} and \frac{x^{2}}{4} have the same denominator, add them by adding their numerators.

$$6.25+\frac{2\times 5x+x^{2}}{4}-\left(2.5-\frac{x}{2}\right)^{2}=\left(\frac{4\left(2.5+\frac{5}{2}\right)}{5}\right)^{2}$$

Do the multiplications in 2\times 5x+x^{2}.

$$6.25+\frac{10x+x^{2}}{4}-\left(2.5-\frac{x}{2}\right)^{2}=\left(\frac{4\left(2.5+\frac{5}{2}\right)}{5}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2.5-\frac{x}{2}\right)^{2}.

$$6.25+\frac{10x+x^{2}}{4}-\left(6.25+5\left(-\frac{x}{2}\right)+\left(-\frac{x}{2}\right)^{2}\right)=\left(\frac{4\left(2.5+\frac{5}{2}\right)}{5}\right)^{2}$$

Express 5\left(-\frac{x}{2}\right) as a single fraction.

$$6.25+\frac{10x+x^{2}}{4}-\left(6.25+\frac{-5x}{2}+\left(-\frac{x}{2}\right)^{2}\right)=\left(\frac{4\left(2.5+\frac{5}{2}\right)}{5}\right)^{2}$$

Calculate -\frac{x}{2} to the power of 2 and get \left(\frac{x}{2}\right)^{2}.

$$6.25+\frac{10x+x^{2}}{4}-\left(6.25+\frac{-5x}{2}+\left(\frac{x}{2}\right)^{2}\right)=\left(\frac{4\left(2.5+\frac{5}{2}\right)}{5}\right)^{2}$$

To find the opposite of 6.25+\frac{-5x}{2}+\left(\frac{x}{2}\right)^{2}, find the opposite of each term.

$$6.25+\frac{10x+x^{2}}{4}-6.25-\frac{-5x}{2}-\left(\frac{x}{2}\right)^{2}=\left(\frac{4\left(2.5+\frac{5}{2}\right)}{5}\right)^{2}$$

Subtract 6.25 from 6.25 to get 0.

$$\frac{10x+x^{2}}{4}-\frac{-5x}{2}-\left(\frac{x}{2}\right)^{2}=\left(\frac{4\left(2.5+\frac{5}{2}\right)}{5}\right)^{2}$$

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

$$\frac{10x+x^{2}}{4}-\frac{-5x}{2}-\frac{x^{2}}{2^{2}}=\left(\frac{4\left(2.5+\frac{5}{2}\right)}{5}\right)^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 4 and 2 is 4. Multiply \frac{-5x}{2} times \frac{2}{2}.

$$\frac{10x+x^{2}}{4}-\frac{2\left(-5\right)x}{4}-\frac{x^{2}}{2^{2}}=\left(\frac{4\left(2.5+\frac{5}{2}\right)}{5}\right)^{2}$$

Since \frac{10x+x^{2}}{4} and \frac{2\left(-5\right)x}{4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{10x+x^{2}-2\left(-5\right)x}{4}-\frac{x^{2}}{2^{2}}=\left(\frac{4\left(2.5+\frac{5}{2}\right)}{5}\right)^{2}$$

Do the multiplications in 10x+x^{2}-2\left(-5\right)x.

$$\frac{10x+x^{2}+10x}{4}-\frac{x^{2}}{2^{2}}=\left(\frac{4\left(2.5+\frac{5}{2}\right)}{5}\right)^{2}$$

Combine like terms in 10x+x^{2}+10x.

$$\frac{20x+x^{2}}{4}-\frac{x^{2}}{2^{2}}=\left(\frac{4\left(2.5+\frac{5}{2}\right)}{5}\right)^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Expand 2^{2}.

$$\frac{20x+x^{2}}{4}-\frac{x^{2}}{4}=\left(\frac{4\left(2.5+\frac{5}{2}\right)}{5}\right)^{2}$$

Since \frac{20x+x^{2}}{4} and \frac{x^{2}}{4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{20x+x^{2}-x^{2}}{4}=\left(\frac{4\left(2.5+\frac{5}{2}\right)}{5}\right)^{2}$$

Combine like terms in 20x+x^{2}-x^{2}.

$$\frac{20x}{4}=\left(\frac{4\left(2.5+\frac{5}{2}\right)}{5}\right)^{2}$$

Add 2.5 and \frac{5}{2} to get 5.

$$\frac{20x}{4}=\left(\frac{4\times 5}{5}\right)^{2}$$

Multiply 4 and 5 to get 20.

$$\frac{20x}{4}=\left(\frac{20}{5}\right)^{2}$$

Divide 20 by 5 to get 4.

$$\frac{20x}{4}=4^{2}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{20x}{4}=16$$

Divide 20x by 4 to get 5x.

$$5x=16$$

Divide both sides by 5.

$$x=\frac{16}{5}$$