Question

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

Answer 1

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`