Solve (x^2+3)^2-7=0

$$( x ^ { 2 } + 3 ) ^ { 2 } - 7 = 0$$

Solve for x (complex solution)

$$x=i\sqrt{3-\sqrt{7}}\approx 0.595187944i$$

$$x=-i\sqrt{3-\sqrt{7}}\approx -0-0.595187944i$$

$$x=-i\sqrt{\sqrt{7}+3}\approx -0-2.376078978i$$

$$x=i\sqrt{\sqrt{7}+3}\approx 2.376078978i$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x^{2}+3\right)^{2}.

$$\left(x^{2}\right)^{2}+6x^{2}+9-7=0$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$x^{4}+6x^{2}+9-7=0$$

Subtract 7 from 9 to get 2.

$$x^{4}+6x^{2}+2=0$$

Substitute t for x^{2}.

$$t^{2}+6t+2=0$$

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 1 for a, 6 for b, and 2 for c in the quadratic formula.

$$t=\frac{-6±\sqrt{6^{2}-4\times 1\times 2}}{2}$$

Do the calculations.

$$t=\frac{-6±2\sqrt{7}}{2}$$

Solve the equation t=\frac{-6±2\sqrt{7}}{2} when ± is plus and when ± is minus.

$$t=\sqrt{7}-3 t=-\sqrt{7}-3$$

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

$$x=-i\sqrt{3-\sqrt{7}} x=i\sqrt{3-\sqrt{7}} x=-i\sqrt{\sqrt{7}+3} x=i\sqrt{\sqrt{7}+3}$$