SOLUTION: find the value or values of p in the quadratic equation p^2+13p-30=0

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Question 333672: find the value or values of p in the quadratic equation p^2+13p-30=0
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

p%5E2%2B13p-30=0 Start with the given equation.


Notice that the quadratic p%5E2%2B13p-30 is in the form of Ap%5E2%2BBp%2BC where A=1, B=13, and C=-30


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


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


p+=+%28-%2813%29+%2B-+sqrt%28+%2813%29%5E2-4%281%29%28-30%29+%29%29%2F%282%281%29%29 Plug in A=1, B=13, and C=-30


p+=+%28-13+%2B-+sqrt%28+169-4%281%29%28-30%29+%29%29%2F%282%281%29%29 Square 13 to get 169.


p+=+%28-13+%2B-+sqrt%28+169--120+%29%29%2F%282%281%29%29 Multiply 4%281%29%28-30%29 to get -120


p+=+%28-13+%2B-+sqrt%28+169%2B120+%29%29%2F%282%281%29%29 Rewrite sqrt%28169--120%29 as sqrt%28169%2B120%29


p+=+%28-13+%2B-+sqrt%28+289+%29%29%2F%282%281%29%29 Add 169 to 120 to get 289


p+=+%28-13+%2B-+sqrt%28+289+%29%29%2F%282%29 Multiply 2 and 1 to get 2.


p+=+%28-13+%2B-+17%29%2F%282%29 Take the square root of 289 to get 17.


p+=+%28-13+%2B+17%29%2F%282%29 or p+=+%28-13+-+17%29%2F%282%29 Break up the expression.


p+=+%284%29%2F%282%29 or p+=++%28-30%29%2F%282%29 Combine like terms.


p+=+2 or p+=+-15 Simplify.


So the solutions are p+=+2 or p+=+-15