Question

x squer -8x-128=0 how can solve

Answer 1

${x}^{2} - 8x - 128 = 0 = >$
${x}^{2} - 16x + 8x - 128 = 0$
$= > x(x - 16) + 8(x - 16) = 0$
=> (x-16)(x+8) =0 => x=16 and -8.