Roots of quadratic equation by completing the square or other method?

Question

I'm trying to find solution(s) to the following equation:

$x^2 - 5x + 3 = 0$

It seems like it can't be factored normally so I tried solving by completing the square:

$x^2-5x=-3$

$x^2-5x+6.25=-0.5$

$(x-2.5)^2 = -0.5$

That's where I get stuck since you can't get the real number square root of a negative number.

Is there another method I could use to solve this quadratic equation? Did I make a mistake?

Answer 1

$(x-2.5)^2+3-2.5^2=0$

$x=(2.5)+(3.25)^{1/2}$

$x=(2.5)-(3.25)^{1/2}$

Also, since you're asking for another method, try the quadratic formula. $x = \frac{-b\pm \sqrt{b^{2} - 4 ac}}{2a}$ where $a, b, c$ are the coefficients of $ax^2+bx+c=0$

Answer 2

$x^2-5x+3$ can be written as $(x-\frac 52)^2-\frac{13}4=0$ $$(x-\frac52)^2=\frac {13}4$$ $$(x-\frac52)=\pm\frac {\sqrt 13}{\sqrt 4}$$ $$(x-\frac52)=\pm\frac {\sqrt 13}{2}$$ $x= \frac {5+\sqrt13}{2}$ or $x=\frac{5-\sqrt 13}{2}$