Question 65134
Solve by completing the square.  Show your steps: 

2x^2 - 3x + 5 = 0  First we'll divide both sides by 2.  Why?  Because in the standard form of the quadratic equation (Ax^2+Bx+C=0). I find the problem easier if A is 1.  Also, if the quadratic is factorable and A is 1 then B is the sum of the products of C.

Dividing by 2 we get:

x^2-(3/2)x+5/2=0 subtract 5/2 from both sides:

x^2-(3/2)x=-5/2  If we take 1/2 of the B term (1/2)(3/2), 
square it ((1/2)(3/2))^2 and add it to both sides, the left side will be a perfect square.

((1/2)(3/2))^2=(3/4)^2=9/16. Now we'll add this to both sides and we have:

x^2-(3/2)x+9/16=-5/2+9/16 =-40/16+9/16 simplifying

x^2-(3/2)x+9/16=-31/16 Factoring the left side we get:

(x-3/4)^2=-31/16  Take sqrt of both sides:

x-3/4=+or-sqrt(-31/16) =+or-((i)sqrt(31)/4  Now add 3/4 to both sides:

x=3/4+or-((i)sqrt(31))/4  so we have:

x=(3+(i)sqrt(31))/4
x=(3-(i)sqrt(31))/4





Hope this helps some.  Happy holidays-----ptaylor