# How do you solve (x + 3) (x + 4) = 5?

Aug 5, 2015

${x}_{1 , 2} = \frac{- 7 \pm \sqrt{21}}{2}$

#### Explanation:

Start by using the FOIL method to expand the two parantheses

$\left(x + 3\right) \left(x + 4\right) = {x}^{2} + 3 x + 4 x + 12 = {x}^{2} + 7 x + 12$

${x}^{2} + 7 x + 12 = 5$

Rearrange so that all the terms are on one side of the equation

${x}^{2} + 7 x + 7 = 0$

${x}_{1 , 2} = \frac{- 7 \pm \sqrt{{7}^{2} - 4 \cdot 1 \cdot 7}}{2 \cdot 1}$

${x}_{1 , 2} = \frac{- 7 \pm \sqrt{21}}{2} = \left\{\begin{matrix}{x}_{1} = \frac{- 7 + \sqrt{21}}{2} \\ {x}_{2} = \frac{- 7 - \sqrt{21}}{2}\end{matrix}\right.$

Check to see if these values are actual solutions

$\left(\frac{- 7 + \sqrt{21}}{2} + 3\right) \left(\frac{- 7 + \sqrt{21}}{2} + 4\right) = 5$

${\left(\frac{- 7 + \sqrt{21}}{2}\right)}^{2} + 7 \cdot \left(\frac{- 7 + \sqrt{21}}{2}\right) + 12 = 5$

$\frac{49 - 14 \sqrt{21} + 21}{4} + \frac{- 49 + 7 \sqrt{21}}{2} = - 7$

$70 - \textcolor{red}{\cancel{\textcolor{b l a c k}{14 \sqrt{21}}}} - 98 + \textcolor{red}{\cancel{\textcolor{b l a c k}{14 \sqrt{21}}}} = - 28$

$- 28 = - 28$ $\to$ ${x}_{1}$ is a valid solution!

Now check for the second solution

${\left(\frac{- 7 - \sqrt{21}}{2}\right)}^{2} + 7 \cdot \left(\frac{- 7 - \sqrt{21}}{2}\right) + 12 = 5$

$\frac{49 + 14 \sqrt{21} + 21}{4} + \frac{- 49 - 7 \sqrt{21}}{2} = - 7$

$70 + \textcolor{red}{\cancel{\textcolor{b l a c k}{14 \sqrt{21}}}} - 98 - \textcolor{red}{\cancel{\textcolor{b l a c k}{14 \sqrt{21}}}} = - 28$

$- 28 = - 28 \to {x}_{2}$ is a valid solution as well!