Respuesta :

6NAI9

[tex] \bold{ANSWER}[/tex]

On solving the following quadratic equation ,

[tex] {x}^{2} - 12x - 64 = 0[/tex]

[tex] {x}^{2} - 16x + 12x - 64 = 0[/tex]

[tex]x(x - 16) + 12(x - 16) = 0[/tex]

[tex](x + 12)(x - 16) = 0[/tex]

so,

x+12 = 0

[tex] \huge \boxed{x = -12}[/tex]

x-16 =0

[tex] \huge \boxed{x = 16}[/tex]

abhaysilver1

Here is your answer

[tex]{x}^{2}-12x-64=0[/tex]

[tex]{x}^{2}-16x+4x-64=0[/tex]

[tex]x(x-16)+4(x-16)=0[/tex]

[tex](x-16)(x+4)=0[/tex]

So,

x-16=0 or x+4=0

[tex]<b>x=16 or x=-4</b>[/tex]

HOPE IT IS USEFUL

Otras preguntas