x2 + x - 6 = 0
x2 + 3x - 2x - 6 = 0
x(x+3) -2(x+3) = 0 so (x+3)(x-2) = 0 so x+3 = 0 or x-2 = 0
so that x = -3 or x = 2
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x2 + 4x + 3 = 0 (x + 1)(x + 3) = 0 x ∈ {-3, -1}
(2-r)e-rr
(4 ± i2) where i2 = -1
x2 + 13x = -30 ∴ x2 + 13x + 30 = 0 ∴ (x + 3)(x + 10) = 0 ∴ x ∈ {-3, -10}
x2 + 7x = 0 => x*(x + 7) = 0 => x = 0 or x + 7 = 0 so that x = 0 or x = -7