SOLUTION: How would I solve this equation? 2y^2 - 3y - 6 = 0

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Question 309947: How would I solve this equation?
2y^2 - 3y - 6 = 0
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

2y%5E2-3y-6=0 Start with the given equation.


Notice that the quadratic 2y%5E2-3y-6 is in the form of Ay%5E2%2BBy%2BC where A=2, B=-3, and C=-6


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


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


y+=+%28-%28-3%29+%2B-+sqrt%28+%28-3%29%5E2-4%282%29%28-6%29+%29%29%2F%282%282%29%29 Plug in A=2, B=-3, and C=-6


y+=+%283+%2B-+sqrt%28+%28-3%29%5E2-4%282%29%28-6%29+%29%29%2F%282%282%29%29 Negate -3 to get 3.


y+=+%283+%2B-+sqrt%28+9-4%282%29%28-6%29+%29%29%2F%282%282%29%29 Square -3 to get 9.


y+=+%283+%2B-+sqrt%28+9--48+%29%29%2F%282%282%29%29 Multiply 4%282%29%28-6%29 to get -48


y+=+%283+%2B-+sqrt%28+9%2B48+%29%29%2F%282%282%29%29 Rewrite sqrt%289--48%29 as sqrt%289%2B48%29


y+=+%283+%2B-+sqrt%28+57+%29%29%2F%282%282%29%29 Add 9 to 48 to get 57


y+=+%283+%2B-+sqrt%28+57+%29%29%2F%284%29 Multiply 2 and 2 to get 4.


y+=+%283%2Bsqrt%2857%29%29%2F%284%29 or y+=+%283-sqrt%2857%29%29%2F%284%29 Break up the expression.


So the solutions are y+=+%283%2Bsqrt%2857%29%29%2F%284%29 or y+=+%283-sqrt%2857%29%29%2F%284%29