Question

12-{6x+[3x+(x-7)(x+7)]-(2x+3)^2}=-2x^2+5[(x+1)^2-3(x+6)]

ayuda por favor

Answer 1

$12-\{6x+[3x+(x-7)(x+7)]-(2x+3)^2\}= -2x^2+5[(x+1)^2- \\ 3(x+6)] \\ \\ \\ 12-\{6x+[3x+(x ^{2}-49)]-(4x^2+12x+9)\}= -2x^2+5[(x+1)^2- \\ 3(x+6)] \\ \\ \\ 12-\{6x+[3x+x ^{2}-49]-4x^2-12x-9\}= -2x^2+5[(x+1)^2- \\ 3(x+6)] \\ \\ \\ 12-\{9x+x ^{2}-49-4x^2-12x-9\}= -2x^2+5[(x+1)^2- \\ 3(x+6)] \\ \\ \\ 12-\{-3x^2-3x-58\}= -2x^2+5[(x+1)^2- 3(x+6)] \\ \\ \\ 12+3x^2+3x+58= -2x^2+5[(x+1)^2- 3(x+6)] \\ \\ \\ 3x^2+3x+70= -2x^2+5[x^2+2x+1- 3x-18] \\ \\ \\ 3x^2+3x+70= -2x^2+5[x^2- x-17]$

$3x^2+3x+70= -2x^2+5x^2-5 x-85 \\ \\ \\ 3x^2+3x+70+2x^2-5x^2+5 x+85=0 \\ \\ \\ 8x+155=0 \\ \\ \\ 8x = -155\qquad \boxed{x= - \frac{155}{8} }$


Espero que te sirva, salu2!!!!