How do you find the zeros, real and imaginary, of $y= 23x^2+18x-24$ using the quadratic formula?

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Answer 1 · 2018-03-27T13:40:32

There are two real roots:

$-9/23-sqrt(633)/23$ and $-9/23+sqrt(633)/23$

We have:

$y = 23x^2+18x-24$

So the roots are given by the solution of the equation:

$23x^2+18x-24 = 0$

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

With $a=23, b=18$ and $c=-24$ we have:

$x = (-18+-sqrt(18^2-4(23)(-24)))/(2(23))$
$\ \ = (-18+-sqrt(324+2208))/(46)$
$\ \ = (-18)/(46)+-sqrt(2532)/(46)$
$\ \ = -9/23+-sqrt(633)/23$

Hence, there are two real roots:

$-9/23-sqrt(633)/23$ and $-9/23+sqrt(633)/23$

How do you find the zeros, real and imaginary, of y= 23x^2+18x-24y= 23x^2+18x-24 using the quadratic formula?