How do you solve by completing the square for $x^{2} - 7 x + 12 = 0$ $x^2-7x+12=0$ ?

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Answer 1 · 2015-05-27T18:38:37

$f (x) = x^{2} - 7 x + \frac{49}{4} - \frac{49}{4} + 12 = 0$ $f(x) = x^2 - 7x + 49/4 - 49/4 + 12 = 0$

$f (x) = {(x - \frac{7}{2})}^{2} - \frac{49}{4} + \frac{48}{4} = 0$ $f(x) = (x - 7/2)^2 - 49/4 + 48/4 = 0$

$(x - \frac{7}{2}) = \pm \frac{1}{2}$ $(x - 7/2) = +- 1/2$

$x = \frac{7}{2} \pm \frac{1}{2}$ $x = 7/2 +- 1/2$

$x = \frac{8}{2} = 4$ $x = 8/2 = 4$
or
$x = \frac{6}{2} = 3$ $x=6/2 =3$

How do you solve by completing the square for x^2-7x+12=0x2−7x+12=0x^2-7x+12=0?