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

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Answer 1 · 2016-04-24T07:28:03

The solutions are:
$y = 2$

$y = 1$

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

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:
$Delta=b^2-4*a*c$
$= (-3)^2-(4* 1 * 2)$
$= 9 - 8 = 1$

The solutions are found using the formula
$y=(-b+-sqrtDelta)/(2*a)$

$y = (-(-3)+-sqrt(1))/(2*1) = (3 +- 1 )/2$

$y = (3 + 1 ) /2 = 4 / 2 = 2$

$y = (3 - 1 ) /2 = 2 / 2 = 1$

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