Not that standard form is y=ax^2+bx+c
and that vertex form is y=color(red)(a)(x+color(red)(b/(2a)))^2+c +k
where k is a correction constant in your case a=1
'~~~~~~~~~~~~~~~~~~~~~~~~~~~
Given:" "y=x^2+3x-9=0
write as: 0=(x^2+3x)-9
Add the correction constant k
0=(x^2+3x)-9+k
Remove the x from 3x
0=(x^2+3)-9+k
Halve the 3
0=(x^2+3/2)-9+k
Move the power from x^2 to outside the bracket
0=(x+3/2)^2-9+k" "color(red)(larr "the error comes from "a(3/(2a))^2)
but k+(3/2)^2=0 => k=-9/4 giving
color(blue)("Vertex form "->0=(x+3/2)^2- 45/4
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Take the -45/4 to the other side of the equals and change its sign.
(x+3/2)^2=+45/4
Square root both sides
x+3/2=sqrt(45/4)
Move the 3/2 to the other side of =
x=-3/2+-sqrt(5xx9)/sqrt(4)
color(blue)(x=-3/2+-(3sqrt(5))/2)
color(blue)(x~~1.854" "and" " -4.854 )
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