Решите уравнение 2x^2-2x-12=(x+2)^2

2х²-2х-12=(х+2)²
2х²-2х-12=х²+4х+4
2х²-2х-12-х²-4х-4=0
х²-6х-16=0
D=36+4*16*1=100
x₁=6+10/2=8
х₂=6-10/2=-2

2х^2 – 2х – 12 = (х + 2)^2
2(х^2 – х – 6) = (х + 2)^2
2(х^2 + 2х – 3х – 6) = (х + 2)^2
2(х • (х + 2) – 3(х + 2)) = (х + 2)^2
2(х – 3) • (х + 2) = (х + 2)^2
2(х – 3) • (х + 2) – (х + 2)^2 = 0
(х + 2) • (2(х – 3) – (х + 2)) = 0
(х + 2) • (2х – 6 – х – 2) = 0
(х + 2) • (х – 8) = 0
х + 2 = 0
х – 8 = 0
х = – 2
х = 8