How do you find the roots, real and imaginary, of #y= (3/4x+1460)x+300 # using the quadratic formula?

Question

1508 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-15T12:14:28

#x=(-2920+-20sqrt(21307))/3#

We take #(3/4x+1460)x+300# and expand it:

#3/4x^2+1460x+300#

We equate this to #0#, and use the quadratic formula to solve it.

The formula states that:

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

Here, #a=3/4,b=1460,c=300#. Inputting:

#x=(-1460+-sqrt(1460^2-4(3/4)(300)))/(2*3/4)#

#x=(-1460+-sqrt(2131600-900))/(3/2)#

#x=(-1460+-sqrt(2130700))/(3/2)#

#x=(-2920+-20sqrt(21307))/3#