How do you find the zeros, real and imaginary, of $y= -7x^2-3x+2$ using the quadratic formula?

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Answer 1 · 2018-01-26T05:49:09

About $-0.79$ and $0.36$ .

This quadratic is in standard form: $ax^2+bx+c$

So for $y=-7x^2-3x+2$ ,
$a=-7$ ,
$b=-3$ , and
$c=2$ .

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

and plug in our $a$ , $b$ , and $c$ values and simplify:

$x=(3+-sqrt((-3)^2-4(-7)(2)))/(2(-7))$

$=(3+-sqrt(9+56))/(-14)$

$=(3+-sqrt(65))/(-14)$

which splits into the two solutions

$x=(3+sqrt(65))/(-14) ~~ -0.79$
and
$x=(3-sqrt(65))/(-14) ~~ 0.36$

How do you find the zeros, real and imaginary, of y= -7x^2-3x+2 y= -7x^2-3x+2 using the quadratic formula?