SOLUTION: Hi! I really need help with this question. I have no idea how to answer it. Please help. It says to: Use the quadratic forumula to solve the following. Leave irrational roots in th

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Hi! I really need help with this question. I have no idea how to answer it. Please help. It says to: Use the quadratic forumula to solve the following. Leave irrational roots in th      Log On


   

Question 131276This question is from textbook structure and method
: Hi! I really need help with this question. I have no idea how to answer it. Please help. It says to: Use the quadratic forumula to solve the following. Leave irrational roots in the simplest form.
3x^2-4x-5=0
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 This question is from textbook structure and method

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Solve, using the quadratic formula:
3x%5E2-4x-5+=+0
The quadratic formula is: x+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F2a, and the a, b, and, of course, are the coefficients of the three terms in the given quadratic equation.
In this case, a = 3, b = -4, and c = -5, so you make the appropriate substitutions into the formula and solve for x.
x+=+%28-%28-4%29%2B-sqrt%28%28-4%29%5E2-4%283%29%28-5%29%29%29%2F2%283%29
x+=+%284%2B-sqrt%2816-%28-60%29%29%29%2F6
x+=+%284%2B-sqrt%2876%29%29%2F6
x+=+%284%2B-sqrt%284%2A19%29%29%2F6
So, x+=+%282%2B-sqrt%2819%29%29%2F3