How do you solve #5t ^ { 2} - 12t - 17= 0#?

1 Answer
Shantelle
Nov 27, 2017

#t = 17/5# (or #3.4#) and #t = -1#

Explanation:

#5t^2 - 12t - 17 = 0#

We solve by factoring.

This equation is written in standard quadratic form, or #ax^2 + bx + c = 0#.

In order to factor, we need to find two numbers that add up to #b# and multiply up to #a*c#.
First, let's list out what #a#, #b#, and #c# are.
#a = 5#
#b = -12#
#c = -17#

So we need 2 numbers that add up to #-12# and multiply up to #-85# (because #5 * -17 = -85#).

So #5# and #-17# work:
#5 - 17 = -12#
#5 * -17 = -85#

Therefore, we now put them back into the equation, like this:
#5t^2 + 5t - 17t - 17 = 0#

Now we factor this by grouping.

#5t^2# and #5t# both have #5t# in common, so let's factor that out. We also know that #-17t# and #-17# have #-17# in common, so let's factor that out too:
#5t(t+1) -17(t+1) = 0#

Now we can combine it like this:
#(5t-17)(t+1) = 0#

Since both of the things in parenthesis multiplied by each other equal to zero, we can set each of the parenthesis equal to zero:
#5t - 17 = 0# and #t + 1 = 0#

Now we solve each of them for #t#.
#5t = 17# and #t = -1#
#t = 17/5# or #3.4# and #t = -1#

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