Question 774694
{{{x^2-15x+56=0}}}....first factor completely so we can use zero product rule; write {{{-15x}}} as {{{-7x-8x}}}

{{{x^2-7x-8x+56 = 0}}}......group

{{{(x^2-7x)-(8x-56 )= 0}}}...factor

{{{x(x-7)-8(x-7 )= 0}}}

{{{(x-8)(x-7) = 0}}}

solutions:

if {{{(x-8) = 0}}}, then {{{x=8}}}

if {{{(x-7) = 0}}}, then {{{x=7}}}

so, you have two real solutions

let's see them on a graph

{{{ graph( 600, 600, -2, 15, -2, 10, x^2-15x+56) }}}