[tex] = > (x - 2)(x + 5) = 0 \\ \\ = > x(x + 5) - 2(x + 5) = 0 \\ \\ = > {x}^{2} + 5x - 2x - 10 = 0 \\ \\ = > {x}^{2} + 3x - 10 = 0 \\ \\ If \: we \: have \: to \: find \: value \: of \: x \: \\ then \: substitute \: both \: the \: term \: \\ equal \: to \: zero \\ \\ = > x - 2 = 0 \: \: \: \: \: \: x + 5 = 0 \\ \\ = > x = 2 \: \: \: \: \: \: x = - 5 \\ \\ = > x = 2 \: or \: - 5[/tex]
