Solve 4-b^4+3b^2-36=0

$$4 - b ^ { 4 } + 3 b ^ { 2 } - 36 = 0$$

Solve for b

$$b=2^{\frac{5}{4}}e^{\frac{\arctan(\frac{\sqrt{119}}{3})i}{2}}\approx 1.891673102+1.441675111i$$

$$b=2^{\frac{5}{4}}e^{\frac{\arctan(\frac{\sqrt{119}}{3})i+2\pi i}{2}}\approx -1.891673102-1.441675111i$$

$$b=2^{\frac{5}{4}}e^{-\frac{\arctan(\frac{\sqrt{119}}{3})i}{2}}\approx 1.891673102-1.441675111i$$

$$b=2^{\frac{5}{4}}e^{\frac{-\arctan(\frac{\sqrt{119}}{3})i+2\pi i}{2}}\approx -1.891673102+1.441675111i$$

Steps Using the Quadratic Formula

Subtract 36 from 4 to get -32.

$$-32-b^{4}+3b^{2}=0$$

Substitute t for b^{2}.

$$-t^{2}+3t-32=0$$

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, 3 for b, and -32 for c in the quadratic formula.

$$t=\frac{-3±\sqrt{3^{2}-4\left(-1\right)\left(-32\right)}}{-2}$$

Do the calculations.

$$t=\frac{-3±\sqrt{-119}}{-2}$$

Solve the equation t=\frac{-3±\sqrt{-119}}{-2} when ± is plus and when ± is minus.

$$t=\frac{-\sqrt{119}i+3}{2} t=\frac{3+\sqrt{119}i}{2}$$

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

$$b=2^{\frac{5}{4}}e^{-\frac{\arctan(\frac{\sqrt{119}}{3})i}{2}} b=2^{\frac{5}{4}}e^{\frac{-\arctan(\frac{\sqrt{119}}{3})i+2\pi i}{2}} b=2^{\frac{5}{4}}e^{\frac{\arctan(\frac{\sqrt{119}}{3})i+2\pi i}{2}} b=2^{\frac{5}{4}}e^{\frac{\arctan(\frac{\sqrt{119}}{3})i}{2}}$$