217. Solve the equation: a) $x^{2}-6 x+13=0$; b) $9 x^{2}+12 x+$ $+29=0$
Solution. a) Let's find the discriminant using the formula $D=b^{2}-4 a c$. Since $a=1, b=-6, c=13$, then $D=(-6)^{2}-4 \cdot 1 \cdot 13=36-52=$ $=-16 ; \sqrt{D}=\sqrt{-16}=\sqrt{16 \cdot(-1)}=4 i$. The roots of the equation are found using the formulas $x_{1}=\frac{-b-\sqrt{D}}{2 a} ; x_{2}=\frac{-b+\sqrt{D}}{2 a}$ :

$$
x_{1}=\frac{6-4 i}{2}=\frac{2(3-2 i)}{2}=3-2 i ; x_{2}=\frac{6+4 i}{2}=\frac{2(3+2 i)}{2}=3+2 i
$$

b) Here $a=9, b=12, c=29$. Therefore, $D^{2}=b^{2}-4 a c=$ $=12^{2}-4 \cdot 9 \cdot 29=144-1044=-900, \quad \sqrt{D}=\sqrt{-900}=\sqrt{900(-1)}=30 i$. We find the roots of the equation:

$$
\begin{aligned}
& x_{1}=\frac{-12-30 i}{18}=\frac{6(-2-5 i)}{18}=\frac{-2-5 i}{3}=-\frac{2}{3}-\frac{5}{3} i \\
& x_{2}=\frac{-12+30 i}{18}=\frac{6(-2+5 i)}{18}=\frac{-2+5 i}{3}=-\frac{2}{3}+\frac{5}{3} i
\end{aligned}
$$

We see that if the discriminant of a quadratic equation is negative, then the quadratic equation has two conjugate complex roots.

218-221. Solve the quadratic equations: