Question

1. Sa se calculeze: log in baza 2 din( x^2+3x-10)=3

Answer 1

   
[tex]\displaystyle \\ \log_{2}(x^2 + 3x -10) = 3 \\ \\ x^2 + 3x -10 = 2^3 \\ \\ x^2 + 3x -10 = 8 \\ \\ x^2 + 3x -10- 8=0 \\ \\ x^2 + 3x -18=0 \\ \\ x_{12}= \frac{-3 \pm \sqrt{3^2 + 4\cdot18} }{2} = \\ \\ = \frac{-3 \pm \sqrt{9 + 72} }{2} = \frac{-3 \pm \sqrt{81} }{2} =\frac{-3 \pm 9}{2} \\ \\ x_1 = \frac{-3 + 9}{2} = \frac{6}{2} = \boxed{3} \\ \\ x_2 = \frac{-3 - 9}{2} = \frac{-12}{2} = \boxed{-6} \\ \\ [/tex]