SOLUTION: 3z^2+13z+12=0 how do you solve this quadratic equation?

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Question 171866: 3z^2+13z+12=0 how do you solve this quadratic equation?
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

3z%5E2%2B13z%2B12=0 Start with the given equation.


Notice we have a quadratic equation in the form of az%5E2%2Bbz%2Bc where a=3, b=13, and c=12


Let's use the quadratic formula to solve for z


z+=+%28-b+%2B-+sqrt%28+b%5E2-4ac+%29%29%2F%282a%29 Start with the quadratic formula


z+=+%28-%2813%29+%2B-+sqrt%28+%2813%29%5E2-4%283%29%2812%29+%29%29%2F%282%283%29%29 Plug in a=3, b=13, and c=12


z+=+%28-13+%2B-+sqrt%28+169-4%283%29%2812%29+%29%29%2F%282%283%29%29 Square 13 to get 169.


z+=+%28-13+%2B-+sqrt%28+169-144+%29%29%2F%282%283%29%29 Multiply 4%283%29%2812%29 to get 144


z+=+%28-13+%2B-+sqrt%28+25+%29%29%2F%282%283%29%29 Subtract 144 from 169 to get 25


z+=+%28-13+%2B-+sqrt%28+25+%29%29%2F%286%29 Multiply 2 and 3 to get 6.


z+=+%28-13+%2B-+5%29%2F%286%29 Take the square root of 25 to get 5.


z+=+%28-13+%2B+5%29%2F%286%29 or z+=+%28-13+-+5%29%2F%286%29 Break up the expression.


z+=+%28-8%29%2F%286%29 or z+=++%28-18%29%2F%286%29 Combine like terms.


z+=+-4%2F3 or z+=+-3 Simplify.


So the answers are z+=+-4%2F3 or z+=+-3