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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?

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    $\begingroup$ 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? $\endgroup$ Commented Nov 24, 2015 at 4:28
  • $\begingroup$ Ah, you're right. I added 6.5 to the left side and only 2.5 to the right side. $\endgroup$
    – spencer.sm
    Commented Nov 24, 2015 at 4:32
  • $\begingroup$ 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. $\endgroup$
    – fleablood
    Commented Nov 24, 2015 at 4:44

2 Answers 2

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

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

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

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 $ax^2+bx+c=0$

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$x^2-5x+3$ 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 $x=\frac{5-\sqrt 13}{2}$

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