4 x 2 + 4 = 12
x = 4
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(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