# How do you solve y^2-3y+2=0?

Apr 24, 2016

The solutions are:
$y = 2$

$y = 1$

#### Explanation:

${y}^{2} - 3 y + 2 = 0$

The equation is of the form color(blue)(ay^2+by+c=0 where:

$a = 1 , b = - 3 , c = 2$

The Discriminant is given by:
$\Delta = {b}^{2} - 4 \cdot a \cdot c$
$= {\left(- 3\right)}^{2} - \left(4 \cdot 1 \cdot 2\right)$
$= 9 - 8 = 1$

The solutions are found using the formula
$y = \frac{- b \pm \sqrt{\Delta}}{2 \cdot a}$

$y = \frac{- \left(- 3\right) \pm \sqrt{1}}{2 \cdot 1} = \frac{3 \pm 1}{2}$

$y = \frac{3 + 1}{2} = \frac{4}{2} = 2$

$y = \frac{3 - 1}{2} = \frac{2}{2} = 1$