We will use the completing-the-square method again. To do so, we first need to divide both sides by the leading coefficient, $2\text{.}$ $$\begin{aligned} 2x^2+12x\amp=-26\\ \divideunder{2x^2}{2}+\divideunder{12x}{2}\amp=\divideunder{-26}{2}\\ x^2+6x\amp=-13 \end{aligned}$$ Now we can add $\left(\frac{b}{2}\right)^2=(3)^2=9$ to both sides to complete the square. $$\begin{aligned} x^2+6x\addright{9}\amp=-13\addright{9}\\ (x+3)^2\amp=-4 \end{aligned}$$ $$\begin{aligned} x+3\amp=-\sqrt{-4}\amp\amp\text{or}\amp x+3\amp=\sqrt{-4}\\ x+3\amp=-\sqrt{4}\cdot\sqrt{-1}\amp\amp\text{or}\amp x+3\amp=\sqrt{4}\cdot\sqrt{-1}\\ x+3\amp=-2i\amp\amp\text{or}\amp x+3\amp=2i\\ x\amp=-3-2i\amp\amp\text{or}\amp x\amp=-3+2i \end{aligned}$$ The solution set is $\{-3-2i, -3+2i\}\text{.}$