What are the roots of the equation x² + 6x + 13 = 0 in simplest a + bi form?

Here is their argument. given the obtuse angle x, we make a quadrilateral abcd with ∠dab = x, and ∠abc = 90◦, and ad = bc. say the perpendicular bisector to dc meets the perpendicular bisector to ab at p. then pa = pb and pc = pd. so the triangles pad and pbc have equal sides and are congruent. thus ∠pad = ∠pbc. but pab is isosceles, hence ∠pab = ∠pba. subtracting, gives x = ∠pad−∠pab = ∠pbc −∠pba = 90◦. this is a preposterous conclusion – just where is the mistake in the "proof" and why does the argument break down there?

Which equation represents a circle with a center at (–3, –5) and a radius of 6 units? (x – 3)2 + (y – 5)2 = 6 (x – 3)2 + (y – 5)2 = 36 (x + 3)2 + (y + 5)2 = 6 (x + 3)2 + (y + 5)2 = 36

3-12. write each answer with a reasonable number of figures. find the absolute uncertainty and percent relative uncertainty for each answer. (a) [12.41 (±0.09) + 4.16 (±0.01)] x 7.068 2 (±0.000 4) =? (b) [3.26 (±0.10) x 8.47 (±0.05)] - 0.18 (±0.06) =? (c) 6.843 (±0.008) x 104 + [2.09 (±0.04)- 1.63 (±0.01)] =?