Решите уравнение 1)(3a+1)^2-10=(a+3)(a-3) 2)10m+(2-m)^2=63-(3-m)^2 3)(x+4)(x+3)+(x-2)(x+2)=2(3x+4)

1) 9a² + 6a + 1 - 10 = a² - 9
9a² - a² + 6a - 9 +9 = 0
8a² + 6a = 0
2a(4a + 3) = 0
2a = 0
a = 0
4a + 3 =0
4a = -3
a = - 3/4

10m + 4 - 4m + m² = 63 - (9 - 6m + m²)
m² + 6m + 4 = 63 - 9 + 6m - m²
m² + m² + 6m - 6m = 54 - 4
2m² = 50
m² = 25
m = 5
m = - 5

x² +7x + 12 + x² - 4 = 6x + 8
2x²  + 7x - 6x + 8 -8 = 0
2x² + x = 0
x( 2x + 1) = 0
x = 0
2x + 1 = 0
2x = - 1
x = - 0,5