How do you solve $x^2-5x+3=0$ ?

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Answer 1 · 2016-04-02T17:38:39

Use the quadratic formula to find roots:

$x = 5/2+-sqrt(13)/2$

This quadratic equation is in the form $ax^2+bx+c = 0$ with $a=1$ , $b=-5$ and $c=3$ .

It has roots given by the quadratic formula:

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

$=(5+-sqrt(5^2-(4*1*3)))/(2*1)$

$=(5+-sqrt(25-12))/2$

$=(5+-sqrt(13))/2$

That is $x = 5/2+sqrt(13)/2$ or $x = 5/2-sqrt(13)/2$

How do you solve x^2-5x+3=0x^2-5x+3=0?