How do you find the roots, real and imaginary, of $y=-15^2 +40x -34$ using the quadratic formula?

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How do you find the roots, real and imaginary, of $y=-15x^2 +40x -34$ using the quadratic formula?

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Answer 1 · 2018-04-19T01:17:05

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

Where:
$a=-15$
$b=40$
$c=-34$

$:.x=(-b+-sqrt(b^2-4ac))/(2a)$
$x=(-(40)+-sqrt((40)^2-4(-15)(-34)))/(2(-15))$
$x=(-40+-sqrt(-440))/(-30)$

And because one cannot square root any negative integers,
the answer obtained for $y=-15x^2+40x-34$ would remain as:
$x=(-40+-sqrt(-440))/(-30)$

By presenting the roots:
$x=(-40+sqrt(-440))/(-30)$ and $x=(-40-sqrt(-440))/(-30)$