Question

determine el valor de x
$\frac{ x^{2} - 4}{x + 3} = -12$

Answer 1

Ok....
$\frac{ x^{2} -4}{x-3} =-12$
$x^{2} -4=-12(x+3)$
$x^{2} -4=-12x-36$
$x^{2} -4+12x+36=0$
$x^{2} +12x + 32=0$
Realizamos aspa simple:
x --------- 8
x --------- 4
(x+8)(x+4)=0
x+8=0 ----- x+4=0
x=-8 --------- x=-4
Espero haberte ayudado : )

Answer 2

${ \frac{x^2-4}{x+3} =-12}\\{x^2-4=-12(x+3)}\\{x^2-4=-12x-36}\\{x^2+12x-4+36=0}\\{x^2+12x+32=0}\\{(x+8)(x+4)=0}\\\\{x+8=0}\\{ x_{1}=-8}\\\\{x+4=0}\\{x_{2}=-4}\\\\{salu2. :)}$