How do you use the quadratic formula to solve #3x^2- 8x -15 =0 #?

1 Answer
Feb 12, 2017

Answer:

#x=3.94" and "-1.27# to 2 decimal places (approx. solution)

Explanation:

This is really worth trying to remember. I did it many years ago by making a point of writing it out for every question I attempted. Repetition is the key.

The standard form is:

#y=ax^2+bx+c" "# where #" "x=(-b+-sqrt(b^2-4ac))/(2a)#
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Given: #3x^2-8x-15=0#

Set #a=3"; "b=-8"; "c=-15#

So by substitution we have:

#x=(+8+-sqrt((-8)^2-4(3)(-15)))/(2(3))#

#x=(8+-sqrt(64+180))/6#

#x=4/3+-sqrt(244)/6#

#x=4/3+-sqrt(2^2xx61)/6#

#x=4/3+-sqrt(61)/3" "# This is an exact solution

#x=3.94" and "-1.27# to 2 decimal places (approx. solution)

Tony B