Question

Resolva a equação: (2x-5)(x-2)+(x+2)(1-x)= - 4

Answer 1

Primeiramente vamos desenvolver a equação:

$(2x-5)(x-2)+(x+2)(1-x)=-4\\\\(2x^2-4x-5x+10)+(x-x^2+2-2x)=-4\\\\2x^2-4x-5x+10+x-x^2+2-2x=-4\\\\x^2-10x+12=-4\\\\x^2-10x+12+4=0\\\\x^2-10x+16=0$

Fatorando o trinômio, temos:

$x^2-10x+16=0\\\\(x-2)(x-8)=0\left\langle\begin{matrix}x'-2=0\longrightarrow{x'}=2\\\\x''-8=0\longrightarrow{x''}=8\end{matrix}\right\\\\\\\\\boxed{S=\{2,\ 8\}}$

Answer 2

(2x-5)(x-2)+(x+2)(1-x)= -4
2x²-4x-5x+10+x-x²+2-2x+4=0
2x²-x²-4x-5x+x-2x+4+10+2=0
x² -10x+16=0
Δ=b²-4ac
Δ=(-10)²-4·1·16
Δ=100 - 64
Δ=36
$x=\frac{-b\pm \sqrt{\Delta}}{2a}\\ \\ x=\frac{-(-10)\pm \sqrt{36}}{2\cdot 1}\\ \\ x=\frac{10\pm \sqrt{36}}{2}\\ \\ x'=\frac{10+\sqrt{6}}{2}\to \boxed{x'=8}\\ \\ x''=\frac{10-\sqrt{6}}{2}\to \boxed{x''=2}$

Espero ter ajudado.