Mathematics, 14.09.2020 15:01 AgentPangolin

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

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$(3x+1)(2x-5)=39 \\
6x^2-15x+2x-5=39 \\
6x^2-13x-5=39 \ \ \ \ \ \ \ \ \ \ |-39 \\
6x^2-13x-44=0 \\
6x^2-24x+11x-44=0 \\
6x(x-4)+11(x-4)=0 \\
(6x+11)(x-4)=0 \\
6x+11=0 \ \lor \ x-4=0 \\
6x=-11 \ \lor \ x=4 \\
x=-\frac{11}{6} \ \lor \ x=4 \\
\boxed{x=-1 \frac{5}{6} \hbox{ or } x=4}$

Answer from: biancaTomas010B

   (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}.

Answer from: Quest

2. root 6 is the right answer
Find x.m just typing to fill the character minimum.

Answer from: Quest

Letter b sorry i’m i’m wrong xx

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(3x+1)(2x-5)=39
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