solve for x

(4/4)x^2-(12/4)x+(61/4)=0

3 answers

I realize i need to do "complete the square" and here is what i got...

(4/4)x^2-(12/4)x+(61/4)=0

x^2-3x+15.25=0

x^2-3x=-15.25

x^2-3x+2.25= -15.25+2.25

(x-1.5)^2=-13

x=1.5 +/- sqrt(-13)

I know i am wrong... help?
Strange equation, I will assume you started with
x^2 - 3x + 15.25 = 0
or
x^2 - 3x + 61/4 = 0
or
4x^2 - 12x + 61 = 0

from: x^2 - 3x + 61/4 = 0
x^2 - 3x = -61/4
x^2 - 3x + 9/4 = -61/4 + 9/4
(x - 3/2)^2 = -13
x - 3/2 = ±√-13
x = 3/2 ± √-13

you had that, but are probably worried about the √-13
If you studied complex numbers you would finish by saying
x = 3/2 ±i√13

if you have not studied complex numbers, your conclusion is that the equation has no real solution.
Thank you so much!