(x+3)(x+4)=x2+7x+12
(x + 12)(x - 6) = 0 x = 6 or -12
x2 + 4x - 9 = 5x + 3 ∴ x2 - x - 12 = 0 ∴ (x + 3)(x - 4) = 0 ∴ x ∈ {-3, 4}
x2-7x+12 = 0 (x-3)(x-4) = 0x=3,4
x2+10x+1 = -12+2x x2+8x+13 = 0 The solution: x = -4 + the square root of 3 or x = -4 - the square root of 3
(x + 3)(x + 4)
x2 + 6x + 9 = 81 x2 + 6x = 72 x2 + 6x - 72 = 0 (x+12)(x-6) = 0 x= -12, 6 (two solutions)
(x+3)(x+4)=x2+7x+12
x2 + 6x + 12 = 0 x2 + 6x + 9 = -3 (x + 3)2 = -3 x + 3 = ± √-3 x = -3 ± i√3
(x + 12)(x - 6) = 0 x = 6 or -12
x2+7x+12=0 x2+7x=-12 x+Sqaure Root of 7x= 2(SQ Root Symbol)3
(x+3)(x+4) = x2+7x+12
x2 + 4x - 9 = 5x + 3 ∴ x2 - x - 12 = 0 ∴ (x + 3)(x - 4) = 0 ∴ x ∈ {-3, 4}
x2-7x+12 = 0 (x-3)(x-4) = 0x=3,4
x2 + 13x + 12 = (x + 1)(x + 12)
x2+10x+1 = -12+2x x2+8x+13 = 0 The solution: x = -4 + the square root of 3 or x = -4 - the square root of 3
x2+8x+12 = 0 When factorized: (x+6)(x+2) = 0 Therefore: x = -6 or x = -2