solve for c 3c^2+19c+30=0?
回答 (5)
2009-11-14 3:10 pm
✔ 最佳答案
3c² + 19c + 30 = 0c² + 19/6c = - 10 + (19/6)
c² + 19/6c = - 360/36 + 361/36
(c + 19/6)² = 1/36
c + 19/6 = ± 1/6
c = 1/6 - 19/6, c = - 18/6, c = - 3
c = - 1/6 - 19/6, c = - 20/6, c = - 10/3
Answer: c = - 3, - 10/3
2009-11-14 10:46 pm
Compare to the standard quadratic formula: ax^2 + bx + c = 0
So in that formula, a = 3, b = 19, c = 30.
Plug into the solution formula x = [-b +- sqrt(b^2 - 4ac)] / 2a
So in that formula, a = 3, b = 19, c = 30.
Plug into the solution formula x = [-b +- sqrt(b^2 - 4ac)] / 2a
2009-11-14 11:24 pm
3c^2 + 19c + 30 = 0
3c^2 + 10c + 9c + 30 = 0
(3c^2 + 10c) + (9c + 30) = 0
c(3c + 10) + 3(3c + 10) = 0
(3c + 10)(c + 3) = 0
3c + 10 = 0
3c = -10
c = -10/3
c + 3 = 0
c = -3
â´ c = -10/3, -3
3c^2 + 10c + 9c + 30 = 0
(3c^2 + 10c) + (9c + 30) = 0
c(3c + 10) + 3(3c + 10) = 0
(3c + 10)(c + 3) = 0
3c + 10 = 0
3c = -10
c = -10/3
c + 3 = 0
c = -3
â´ c = -10/3, -3
2009-11-14 10:50 pm
Factor the quadratic:
3c^2+19c+30 = (3c+10)(c+3).
Then setting to zero gives 3c+10 = 0 or c+3 = 0.
So the solutions are c = -10/3 and c = -3.
3c^2+19c+30 = (3c+10)(c+3).
Then setting to zero gives 3c+10 = 0 or c+3 = 0.
So the solutions are c = -10/3 and c = -3.
2009-11-14 10:48 pm
c is 10/3 or -3
收錄日期: 2021-05-01 12:54:07
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