# How do you solve (s+1)^2/10-12/5=15/2?

May 8, 2017

$s = - 1 \pm 3 \sqrt{11}$

#### Explanation:

Given:

$\frac{{\left(s + 1\right)}^{2}}{10} - \frac{12}{5} = \frac{15}{2}$

We want to isolate the ${\left(s + 1\right)}^{2}$ term in order to solve for s

First add $\frac{12}{5}$ to both sides:

$\frac{{\left(s + 1\right)}^{2}}{10} - \frac{12}{5} \textcolor{red}{+ \frac{12}{5}} = \frac{15}{2} \textcolor{red}{+ \frac{12}{5}}$

On the left hand side, the $- \frac{12}{5}$ and $+ \frac{12}{5}$ add to $0$:

$\frac{{\left(s + 1\right)}^{2}}{10} \cancel{- \frac{12}{5}} \cancel{\textcolor{red}{+ \frac{12}{5}}} = \frac{15}{2} \textcolor{red}{+ \frac{12}{5}}$

And on the right hand side, we need to find a common denominator to add the fractions:

$\frac{{\left(s + 1\right)}^{2}}{10} = \frac{15}{2} \textcolor{red}{\left(\frac{5}{5}\right)} + \frac{12}{5} \textcolor{red}{\left(\frac{2}{2}\right)}$

$\frac{{\left(s + 1\right)}^{2}}{10} = \textcolor{red}{\frac{75}{10}} + \textcolor{red}{\frac{24}{10}} = \frac{99}{10}$

Next we multiple both sides by 10 to isolate the ${\left(s + 1\right)}^{2}$ term

$\textcolor{red}{10} \frac{{\left(s + 1\right)}^{2}}{10} = \textcolor{red}{10} \frac{99}{10}$

Which gives

${\left(s + 1\right)}^{2} = 99$

Square root both sides of the equation

$\sqrt{{\left(s + 1\right)}^{2}} = \sqrt{99}$

Which gives

$s + 1 = \pm \sqrt{99}$

NOTE: When taking the square root of any term, the answer can be both negative (-) and positive (+). When you raise a negative number to the second (2) power, the negative multiplies out to give a positive answer.

Now, simplifying the answer, and subtracting 1 from both sides

$s + 1 \textcolor{red}{- 1} = \pm \sqrt{\left(9\right) \left(11\right)} \textcolor{red}{- 1}$

9 is a perfect square, with a square root value of 3, so the answer can be simplified to

$s = - 1 \pm 3 \sqrt{11}$