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

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?

• I think you made a mistake going from $x^2-5x = -3$ to $x^2 - 5x + 6.25 = -0.5$. What did you add to both sides? Commented Nov 24, 2015 at 4:28
• Ah, you're right. I added 6.5 to the left side and only 2.5 to the right side. Commented Nov 24, 2015 at 4:32
• Not all quadratics are solvable in the real numbers, by the way. Example: $x^2 - 5x + 7 = 0$ will get you $(x - 2.5)^2 = -0.75$ which has no real number solution. Commented Nov 24, 2015 at 4:44

$(x-2.5)^2+3-2.5^2=0$
$x=(2.5)+(3.25)^{1/2}$
$x=(2.5)-(3.25)^{1/2}$
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 $ax^2+bx+c=0$
$$x^2-5x+3$$ 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 $$x=\frac{5-\sqrt 13}{2}$$