# How do you solve (x-3)^2 = 5 using the quadratic formula?

May 9, 2018

${\left(x - 3\right)}^{2} = 5$

${x}_{1} = 3 + \setminus \sqrt{5}$
${x}_{2} = 3 - \setminus \sqrt{5}$

#### Explanation:

${\left(x - 3\right)}^{2} = 5$
${x}^{2} - 6 \cdot x + 9 = 5$
${x}^{2} - 6 \cdot x + 9 - 5 = 0$
${x}^{2} - 6 \cdot x + 4 = 0$

$a \cdot {x}^{2} + b \cdot x + c = 0$

${x}_{1 , 2} = \frac{- b \pm \setminus \sqrt{{b}^{2} - 4 \cdot a \cdot c}}{2 \cdot a}$

$a = 1$
$b = 6$
$c = 4$

${x}_{1 , 2} = \frac{- \left(- 6\right) \pm \setminus \sqrt{{6}^{2} - 4 \cdot 1 \cdot 4}}{2 \cdot 1} =$
$= \frac{6 \pm \setminus \sqrt{36 - 16}}{2 \cdot 1} =$
$= \frac{6 \pm \setminus \sqrt{20}}{2} =$
$= \frac{6 \pm \setminus \sqrt{4 \cdot 5}}{2} =$
$= \frac{6 \pm 2 \cdot \setminus \sqrt{5}}{2} =$
$= 3 \pm \setminus \sqrt{5}$

${x}_{1} = 3 + \setminus \sqrt{5}$
${x}_{2} = 3 - \setminus \sqrt{5}$