Question

(3x+1)^2+(x-2)(x+1)=-1 se possivel em bhascara pfv

Answer 1

$Boa \: \: noite!!! \\ \\ {(3x + 1)}^{2} + (x - 2)(x + 1) = - 1 \\ 9 {x}^{2} + 6x + 1 + {x}^{2} + x - 2x - 2 = - 1 \\ 10 {x}^{2} + 5x - 1 = - 1 \\ 10 {x}^{2} + 5x = 0 \\ 5x ^{ \div 5} (2x + 1) = 0 \\ x(2x + 1) = 0 \\ x = 0 \\ 2x + 1 = 0 = > x = - \frac{1}{2} \\ {x}^{1} = - \frac{1}{2} \: e \: {x}^{2} = 0 \\ \\ Espero \: \: ter \: \: ajudado!:)$