Use similarity in figures to solve for the value of the x

Here ${\displaystyle DE\parallel AC}$
From the figure
${\displaystyle \frac{BE}{BC}=\frac{DE}{AC}}$
${\displaystyle \Rightarrow \frac{6}{6+x}=\frac{x-1}{2x+4}}$
${\displaystyle \Rightarrow 6(2x+4)=(6+x)(x-1)}$
${\displaystyle \Rightarrow 12x+24=6x-6+x^2-x}$
${\displaystyle \Rightarrow 12x+24=5x+x^2-6}$
${\displaystyle \Rightarrow x^2+5x-6-12x-24=0}$
${\displaystyle \Rightarrow x^2-7x-30=0}$
${\displaystyle \Rightarrow x^2-10x+3x-30=0}$
${\displaystyle \Rightarrow x(x-10)+3(x-10)=0}$
${\displaystyle \Rightarrow (x-10)(x+3)=0}$
${\displaystyle \Rightarrow }$ either x-10=0 or, x+3=0
${\displaystyle \Rightarrow x=10\Rightarrow x=-3}$
x=-3 is not possitive since length is non negative
Hence x=10