SOLUTION: How do you solve the quadratic equation 3k^2 + 7K=4?

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Question 472582: How do you solve the quadratic equation 3k^2 + 7K=4?
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

3k%5E2%2B7k=4 Start with the given equation.


3k%5E2%2B7k-4=0 Subtract 4 from both sides.


Notice that the quadratic 3k%5E2%2B7k-4 is in the form of Ak%5E2%2BBk%2BC where A=3, B=7, and C=-4


Let's use the quadratic formula to solve for "k":


k+=+%28-B+%2B-+sqrt%28+B%5E2-4AC+%29%29%2F%282A%29 Start with the quadratic formula


k+=+%28-%287%29+%2B-+sqrt%28+%287%29%5E2-4%283%29%28-4%29+%29%29%2F%282%283%29%29 Plug in A=3, B=7, and C=-4


k+=+%28-7+%2B-+sqrt%28+49-4%283%29%28-4%29+%29%29%2F%282%283%29%29 Square 7 to get 49.


k+=+%28-7+%2B-+sqrt%28+49--48+%29%29%2F%282%283%29%29 Multiply 4%283%29%28-4%29 to get -48


k+=+%28-7+%2B-+sqrt%28+49%2B48+%29%29%2F%282%283%29%29 Rewrite sqrt%2849--48%29 as sqrt%2849%2B48%29


k+=+%28-7+%2B-+sqrt%28+97+%29%29%2F%282%283%29%29 Add 49 to 48 to get 97


k+=+%28-7+%2B-+sqrt%28+97+%29%29%2F%286%29 Multiply 2 and 3 to get 6.


k+=+%28-7%2Bsqrt%2897%29%29%2F%286%29 or k+=+%28-7-sqrt%2897%29%29%2F%286%29 Break up the expression.


So the solutions are k+=+%28-7%2Bsqrt%2897%29%29%2F%286%29 or k+=+%28-7-sqrt%2897%29%29%2F%286%29