SOLUTION: Please help me to solve this equation: solve 3x^2+13x+14=0

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Question 475091: Please help me to solve this equation:
solve
3x^2+13x+14=0
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
For more help, check out this quadratic formula solver.

Solved by pluggable solver: Quadratic Formula
Let's use the quadratic formula to solve for x:


Starting with the general quadratic


ax%5E2%2Bbx%2Bc=0


the general solution using the quadratic equation is:


x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29




So lets solve 3%2Ax%5E2%2B13%2Ax%2B14=0 ( notice a=3, b=13, and c=14)





x+=+%28-13+%2B-+sqrt%28+%2813%29%5E2-4%2A3%2A14+%29%29%2F%282%2A3%29 Plug in a=3, b=13, and c=14




x+=+%28-13+%2B-+sqrt%28+169-4%2A3%2A14+%29%29%2F%282%2A3%29 Square 13 to get 169




x+=+%28-13+%2B-+sqrt%28+169%2B-168+%29%29%2F%282%2A3%29 Multiply -4%2A14%2A3 to get -168




x+=+%28-13+%2B-+sqrt%28+1+%29%29%2F%282%2A3%29 Combine like terms in the radicand (everything under the square root)




x+=+%28-13+%2B-+1%29%2F%282%2A3%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)




x+=+%28-13+%2B-+1%29%2F6 Multiply 2 and 3 to get 6


So now the expression breaks down into two parts


x+=+%28-13+%2B+1%29%2F6 or x+=+%28-13+-+1%29%2F6


Lets look at the first part:


x=%28-13+%2B+1%29%2F6


x=-12%2F6 Add the terms in the numerator

x=-2 Divide


So one answer is

x=-2




Now lets look at the second part:


x=%28-13+-+1%29%2F6


x=-14%2F6 Subtract the terms in the numerator

x=-7%2F3 Divide


So another answer is

x=-7%2F3


So our solutions are:

x=-2 or x=-7%2F3




For more help, check out this quadratic formula solver.