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

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

1 Answer

Answer:

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

Explanation:

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)#