How do you find the discriminant and how many and what type of solutions does $y = –2x^2 + x – 3$ have?

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Answer 1 · 2015-05-12T04:40:14

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

$a=-2, b=1, c=-3$

The Disciminant is given by :
$Delta=b^2-4*a*c$
$= (1)^2-(4*(-2)*(-3))$
$= 1-24=-23$

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 $Delta = -23$ , this equation has NO REAL SOLUTIONS

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

As $Delta = -23$ , $x = (-(1)+-sqrt(-23))/(2*-2) = (-(1)+-sqrt(-23))/-4$

How do you find the discriminant and how many and what type of solutions does y = –2x^2 + x – 3y = –2x^2 + x – 3 have?