Question 138827
4y^2-12y+13=0

4y^2 - 12y = -13

4(y^2 - 3y + ?) = -13 + 4*?

4(y^2-3y+(3/2)^2) = -13 + 4*(3/2)^2

4(y - (3/2))^2 = -13 +9
Divide both sides by 4 to get:

(y-(3/2))^2 = -1
Take the square root of both sides to get:

y-(3/2) = +/- i
 
y = [(3/2) + i] or y = [(3/2) - i]


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Cheers,
Stan H.