2007-05-14 02:29:25 · 7 answers · asked by noggle4 2 in Science & Mathematics ➔ Mathematics
(b+2) (b-5)=2b+14
b^2 + 2b - 5b - 10 = 2b + 14
b^2 - 3b - 10 = 2b + 14
b^2 - 5b - 24 = 0
(b - 8)(b + 3) = 0
so...
b = 8 and -3
2007-05-14 02:33:24 · answer #1 · answered by Mathematica 7 · 0⤊ 0⤋
Multiply the left-hand side out. Combine like terms. Then refactor to solve.
b^2 - 3b - 10 = 2b + 14
b^2 - 5b - 24 = 0
(b - 8)(b + 3) = 0
b = 8, -3
2007-05-14 09:33:48 · answer #2 · answered by Anonymous · 0⤊ 0⤋
b^2 - 5b + 2b - 10 = 2b + 14
b^2 - 5b -24 = 0
completing the square:
b^2 - 5b + (5/2)^2 - (5/2)^2 - 24 =0
(b - 5/2)^2 - (25/4 + 24) = 0
(b - 5/2)^2 - (121/4) = 0
(b - 5/2 - 11/2)(b - 5/2 + 11/2) = 0
(b - 8)(b + 3) = 0
b=8
b=-3
2007-05-14 10:06:45 · answer #3 · answered by Anonymous · 0⤊ 0⤋
b² - 3b - 10 = 2b + 14
b² - 5b - 24 = 0
(b - 8).(b + 3) = 0
b = 8, b = - 3
2007-05-14 10:37:43 · answer #4 · answered by Como 7 · 0⤊ 0⤋
b^2 - 3b - 10 = 2b + 14
b^2 - 5b = 24
b^2 - 5b - 24 = 0
(b - 8) (b + 3) = 0
b = 8 or -3
2007-05-14 09:33:20 · answer #5 · answered by Doctor Q 6 · 0⤊ 0⤋
b(b-5)+2(b-5)=2b+14
b^2-5b+2b-10=2b+14
b^2-3b-10=2b+14
b^2-3b-10-2b-14=0
b^2-5b-24=0
b^2-8b+3b-24=0
b(b-8)+3(b-8)=0
(b-8)=0 , (b+3)=0
b=8 or b= -3
2007-05-14 09:42:11 · answer #6 · answered by Wyk123 4 · 0⤊ 0⤋
b= -3, 8
2007-05-14 09:39:35 · answer #7 · answered by Phat MD 4 · 0⤊ 0⤋