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