$ a)2x + 50 = 0 $
$⇒2x = -50$
$⇒x = -25$
$b)(x-2)(x+1)=0$
$⇒$\(\left[ \begin{array}{l}x-2=0\\x+1=0\end{array} \right.\)
$⇒$\(\left[ \begin{array}{l}x=2\\x=-1\end{array} \right.\)
$c)5x(x – 1) = 1 – x$
$⇒5x(x – 1) - (1 - x)=0$
$⇒5x(x – 1) + (x - 1)=0$
$⇒(x - 1)(5x + 1)=0$
$⇒$\(\left[ \begin{array}{l}x - 1=0\\5x + 1=0\end{array} \right.\)
$⇒$\(\left[ \begin{array}{l}x=1\\5x=-1\end{array} \right.\)
$⇒$\(\left[ \begin{array}{l}x=1\\x=\frac{-1}{5}\end{array} \right.\)
$d)(3x – 4)² – (x + 1)² = 0$
$⇒(3x-4-x-1)(3x-4+x+1)=0$
$⇒(2x-5)(4x-3)=0$
$⇒$\(\left[ \begin{array}{l}2x-5=0\\4x-3=0\end{array} \right.\)
$⇒$\(\left[ \begin{array}{l}2x=5\\4x=3\end{array} \right.\)
$⇒$\(\left[ \begin{array}{l}x=\frac{5}{2}\\x=\frac{3}{4}\end{array} \right.\)
$e)x² + 7x + 12 =0$
$⇒x² + 3x + 4x + 12 =0$
$⇒(x² + 3x)+(4x + 12) = 0$
$⇒x(x+3)+4(x+3)=0$
$⇒(x+4)(x+3)=0$
$⇒$\(\left[ \begin{array}{l}x+4=0\\x+3=0\end{array} \right.\)
$⇒$\(\left[ \begin{array}{l}x=-4\\x=-3\end{array} \right.\)
