How do you solve #-0.12x^2 + 1.4x - 3.5 = 0.5# using the quadratic formula?

1 Answer
Jul 9, 2015

Answer:

I found:
#x_1=5#
#x_2=20/3#

Explanation:

Ok...I`ll try to "simplify" it a bit to avoid decimals...(I do not like them!).
I can write it as:
#-12/100x^2+14/10x-35/10=5/10#
getting rid of the denominators (after taking #100# as common) I got:
#-12x^2+140x-400=0#
that is in the form:
#ax^2+bx+c=0#
With:
#a=-12#
#b=140#
#c=-400#
I now use the Quadratic Formula:
#x_(1,2)=(-b+-sqrt(b^2-4ac))/(2a)=(-140+-sqrt(140^2-4(-12*-400)))/(2*-12)=#
#=(-140+-sqrt(400))/-24=(-140+-20)/-24=#
You get two solutions:
#x_1=(-140+20)/-24=5#
#x_2=(-140-20)/-24=160/24=20/3#