Question

(х+1)^2-36=0
Срочноооо

Answer 1

(х + 1)^2 - 36 = 0

х^2 + 2х + 1 - 36 = 0

х^2 + 2х - 35 = 0

k = 1/2b

D = k^2 - ac

D = 1^2 - 1 × (-35) = 36

x = (-k ± √D)/a

x = (-1 ± √36)/1 = -1 ± 6 = - 7; 5

Answer 2

$(x+1)^2-36=0\\\\(x+1-6)(x+1+6)=0\\\\(x-5)(x+7)=0\\\\x-5=0 \quad\quad x+7=0\\\\\boxed{x=5}\quad\quad \boxed{x=-7}$

Воспользовался формулой разности квадратов:  $a^2-b^2=(a-b)(a+b)$