SOLUTION: I'm supposed to use the quadratic formula to solve, but leave irrational roots in the simplest radical form. 3x^2-4x-5=0 Thaks alot

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Question 88292This question is from textbook Algebra structure and method
: I'm supposed to use the quadratic formula to solve, but leave irrational roots in the simplest radical form.
3x^2-4x-5=0
Thaks alot This question is from textbook Algebra structure and method

Answer by tutor_paul(519) About Me  (Show Source):
You can put this solution on YOUR website!
3x%5E2-4x-5=0
Quadratic Formula:
x=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F2a
In this case, a=3, b=-4, c=-5
So, plug and chug:
x=%284%2B-sqrt%28%28-4%29%5E2-%284%2A3%2A%28-5%29%29%29%29%2F%282%2A3%29
Simplify:
x=%284%2B-sqrt%2816%2B60%29%29%2F6
x=%284%2B-sqrt%2876%29%29%2F6
Simplify the radical:
x=(4+-2sqrt(19))/6
Cancel out a 2 from numerator and denominator:
highlight%28x=%282%2B-sqrt%2819%29%29%2F3%29
Good Luck,
tutor_paul@yahoo.com