How do you find the roots, real and imaginary, of #y=(x-2)(-5x+1)-x# using the quadratic formula?

1 Answer
Jun 5, 2016

Answer:

#x=1+-sqrt(60)/10#

Explanation:

First convert the given equation into standard form:

#y=(x-2)(-5x+1)-x#

#color(white)("XXX")rArr y=-5x^2+10x-2#

For the general standard quadratic:
#color(white)("XXX")y=ax^2+bx+c#
the roots are given by the quadratic formula
#color(white)("XXX")x=(-b+-sqrt(b^2-4ac))/(2a)#

In this case
#color(white)("XXX")x=(-10+-sqrt(10^2-4(-5)(-2)))/(2(-5))#

#color(white)("XXX")x=1+-sqrt(60)/10#