# How do you solve y^2-6y+8=0?

Mar 14, 2016

Factor the left side to find
$\textcolor{w h i t e}{\text{XXX}} y = 2 \mathmr{and} y = 4$

#### Explanation:

Given
$\textcolor{w h i t e}{\text{XXX}} {y}^{2} - 6 y + 8 = 0$

Factoring the left side expression
$\textcolor{w h i t e}{\text{XXX}} \left(y - 4\right) \left(y - 2\right) = 0$

which implies
either
$\textcolor{w h i t e}{\text{XXX}} \left(y - 4\right) = 0 \rightarrow y = 4$
or
$\textcolor{w h i t e}{\text{XXX}} \left(y - 2\right) = 0 \rightarrow y = 2$

Mar 14, 2016

$\left\{2 , 4\right\}$

#### Explanation:

${y}^{2} - 6 y + 8 = 0$
$\left(y - 2\right) \left(y - 4\right) = 0$
$\left(y - 2\right) = 0 \implies y = 2$
$\left(y - 4\right) = 0 \implies y = 4$
$\left\{2 , 4\right\}$

Mar 14, 2016

$y = 2$ or $y = 4$

#### Explanation:

Method of factoring:

${y}^{2} - 6 y + 8 = 0$
$\left(y - 2\right) \left(y - 4\right) = 0$
$y - 2 = 0$ or $y - 4 = 0$

Solving for $y$, gives $y = 2$ or $y = 4$

$y = \frac{6 \pm \sqrt{36 - 4 \left(1\right) \left(8\right)}}{2 \left(1\right)} = \frac{6 \pm \sqrt{4}}{2}$
$y = 2$ or $y = 4$