Question

(x+1)²-(2x+3)²=0 forma reduzida

Answer 1

(x2+2x+1)-(4x2+12x+9
x2+2x+1-4x2-12x-9
3x2- 10x-8

Answer 2

Olá, boa noite ☺

Resolução:

$(x+1)^2-(2x+3)^2=0 \\ (x^2+2x+1)-(4x^2+12x+9)=0 \\ x^2+2x+1-4x^2-12x-9=0 \\ \\ x^2-4x^2+2x-12x + 1 - 9 =0 \\ \\ -3x^2-10x-8=0 \\ \\ \texttt{Coeficientes:} \\ \\ a=-3 \ \ b=-10 \ \ c=-8 \\ \\ \Delta= b^{2} -4.a.c \\ \Delta= (-10)^2-4.(-3).(-8) \\ \Delta=100 -96 \\ \Delta=4$

$\texttt{Valores de x:} \\ \\ x= \frac{-b\pm \sqrt{\Delta} }{2.a} \\ \\ x= \frac{-(-10)\pm \sqrt{4} }{2.(-3)} \\ \\ x=\frac{10)\pm 2 }{-6} \\ \\ x'= \frac{10 + 2 }{-6}= \frac{12}{-6} =-2 \\ \\ x''= \frac{10-2}{-6}= \frac{8}{-6} \div \frac{2}{2} = -\frac{4}{3} \\ \\ S=\left \{ -2,-\frac{4}{3} \right \}$

Bons estudos :)