Can someone help me solve c^2 + 12c + 32 = 0?

c^2 + 12c + 32 = 0 Or (c + 8)(c + 4) = 0 Or c = --8 or c = --4.

c^2 + 12c + 32 = 0

(c + 4)(c + 8) = 0

c = -4 and c = -8

(c + 8)(c + 4) = 0
c = - 8 , c = - 4

c^2+12c+32=0
c^2+8c+4c+32=0
c(c+8)+4(c+8)=0
(c+8)(c+4)=0
c=--8 and c=--4

Answer is (c+8)(c+4) = 0

c^2 + 12c + 32 = 0
c^2 + 8c + 4c + 32 = 0
(c^2 + 8c) + (4c + 32) = 0
c(c + 8) + 4(c + 8) = 0
(c + 8)(c + 4) = 0

c + 8 = 0
c = 8

c + 4 =0
c = -4

∴ c = -8 , -4

It factors:
(c + 4)(c + 8) = 0
c + 4 = 0 --OR-- c + 8 = 0
c = -4, -8

x=10
x=14

i used the quadratic equation