# solve the equation: f(x) = x^2 - 3x + 4

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f(x) = x^2 - 3x + 4

Let us use the fomula to solve:

we note that :

delta = b^2 - 4ac = -7 is negative.

If delta < 0 , then the roots are complex ( not real).

Then we have complex roots.

x= [-b + sqrt(b^2 - 4ac)]/2a

x1= [3 + sqrt(9 - 16)/2 = (3+ sqrt(-7) /2 = [3 + (sqrt7)i]/2

x2= (3- (sqrt7)*i]/2

Then , the solution is:

** x= { 3/2 + (sqrt7/2)*i , 3/2 - (sqrt7/2)*i }**

To solve the equation x^2-3x+4 = 0.

We add (3/2) to x^2-3x to make it perfect square and subtract (3/2)^2.

x^2 -3x +(3/2)^2 - (3/2)^2+4 = 0

(x-3/2)^2 - (9/4)+16

(x-3/2)^2 = (-16+9)/4 = -7/4

We take the square root;

x-3/2 = +sqrt(-7/4) or x = -sqrt(-7/4)

x = {3 +sqrt(-7)}/2 or x = {3-sqrt(-7)}/2. Both are complex roots.