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(3x+1)(2x-5)=39
i need help

2 Answers

7 votes
(3x + 1)(2x - 5) = 39
3x(2x - 5) + 1(2x - 5) = 39
3x(2x) + 3x(-5) + 1(2x) + 1(-5) = 39
6x² + (-15x) + 2x + (-5) = 39
6x² - 15x + 2x - 5 = 39
6x² - 13x - 5 = 39
- 39 - 39
6x² - 13x - 44 = 0
x = -(-13) ± √((-13)² - 4(6)(-44))
2(6)
x = 13 ± √(169 + 1056)
12
x = 13 ± √(1225)
12
x = 13 ± 35
12
x = 13 + 35 ∨ x = 13 - 35
12 12
x = 48 ∨ x = -22
12 12
x = 4 ∨ x = -1⁵/₆
The solution set us equal to {4, -1⁵/₆}.
User TheFiddlerWins
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(3x+1)(2x-5)=39 \\ 6x^2-15x+2x-5=39 \\ 6x^2-13x-5=39 \ \ \ \ \ \ \ \ \ \ |-39 \\ 6x^2-13x-44=0 \\ 6x^2-24x+11x-44=0 \\ 6x(x-4)+11(x-4)=0 \\ (6x+11)(x-4)=0 \\ 6x+11=0 \ \lor \ x-4=0 \\ 6x=-11 \ \lor \ x=4 \\ x=-(11)/(6) \ \lor \ x=4 \\ \boxed{x=-1 (5)/(6) \hbox{ or } x=4}
User Kenji Yoshida
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8.4k points

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