How do you solve $x^2 - 12x + 52 = 0$ ?

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Answer 1 · 2015-06-08T12:05:36

$x^2 - 12x + 52 = 0$

The equation is of the form $color(blue)(ax^2+bx+c=0$ where:
$a=1, b=-12, c=52$

The Discriminant is given by:
$Delta=b^2-4*a*c$
$= (-12)^2-(4*1*52)$
$= 144 - 208$
$= -64$

If $Delta=0$ then there is only one solution.
(for $Delta>0$ there are two solutions,
for $Delta<0$ there are no real solutions)

As $Delta = -64$ , this equation has NO REAL SOLUTIONS

Note :
The solutions are normally found using the formula
$x=(-b+-sqrtDelta)/(2*a)$

As $Delta = -64$ , $x = (-(-12)+-sqrt(-64))/(2*1) = (12+-sqrt(-64))/2$

Answer 2 · 2015-06-08T12:06:40

$x^2-12x+52$ is of the form $ax^2+bx+c$ with $a=1$ , $b=-12$ and $c=52$ . This has discriminant given by the formula:

$Delta = b^2-4ac = (-12)^2-(4xx1xx52)$

$=144-208=-64 < 0$

So $x^2-12x+52=0$ has no real roots. It has two distinct complex roots.

How do you solve x^2 - 12x + 52 = 0x^2 - 12x + 52 = 0?