How do you find the discriminant and how many solutions does $x^{2} - 11 x + 28 = 0$ $x^2 -11x + 28 = 0$ have?

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Answer 1 · 2015-05-05T07:42:44

The equation is of the form $a x^{2} + b x + c = 0$ $color(blue)(ax^2+bx+c=0$ where:

$a = 1, b = - 11, c = 28$ $a=1, b=-11, c=28$

The Disciminant is given by :
$Delta=b^2-4*a*c$
$= (-11)^2-(4*1*28)$
$= 121-112=9$

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 $Delta = 9$ , this equation has TWO REAL SOLUTIONS

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

As $Delta = 9$ , $x = (-(-11)+-sqrt9)/(2*1) = (11+-3)/(2*1) = 14/2 or 8/2 = 7 or 4$

$x = 4,7$ are the two solutions

How do you find the discriminant and how many solutions does x^2 -11x + 28 = 0x2−11x+28=0x^2 -11x + 28 = 0 have?