Solve 2x^4+2x^2+3=0

$$2 x ^ { 4 } + 2 x ^ { 2 } + 3 = 0$$

Solve for x (complex solution)

$$x=\frac{\sqrt[4]{24}e^{\frac{\left(\arctan(\sqrt{5})+\pi \right)i}{2}}}{2}\approx -0.601973783+0.9286401i$$

$$x=\frac{\sqrt[4]{24}e^{\frac{\arctan(\sqrt{5})i+3\pi i}{2}}}{2}\approx 0.601973783-0.9286401i$$

$$x=\frac{\sqrt[4]{24}e^{\frac{-\arctan(\sqrt{5})i+3\pi i}{2}}}{2}\approx -0.601973783-0.9286401i$$

$$x=\frac{\sqrt[4]{24}e^{\frac{-\arctan(\sqrt{5})i+\pi i}{2}}}{2}\approx 0.601973783+0.9286401i$$

Steps Using the Quadratic Formula

Substitute t for x^{2}.

$$2t^{2}+2t+3=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 2 for a, 2 for b, and 3 for c in the quadratic formula.

$$t=\frac{-2±\sqrt{2^{2}-4\times 2\times 3}}{2\times 2}$$

Do the calculations.

$$t=\frac{-2±\sqrt{-20}}{4}$$

Solve the equation t=\frac{-2±\sqrt{-20}}{4} when ± is plus and when ± is minus.

$$t=\frac{-1+\sqrt{5}i}{2} t=\frac{-\sqrt{5}i-1}{2}$$

Since x=t^{2}, the solutions are obtained by evaluating x=±\sqrt{t} for each t.

$$x=\frac{\sqrt[4]{24}e^{\frac{-\arctan(\sqrt{5})i+3\pi i}{2}}}{2} x=\frac{\sqrt[4]{24}e^{\frac{-\arctan(\sqrt{5})i+\pi i}{2}}}{2} x=\frac{\sqrt[4]{24}e^{\frac{\arctan(\sqrt{5})i+3\pi i}{2}}}{2} x=\frac{\sqrt[4]{24}e^{\frac{\left(\arctan(\sqrt{5})+\pi \right)i}{2}}}{2}$$