Solve b^4-29b^2-96=0

$${ b }^{ 4 } -29 { b }^{ 2 } -96 = 0$$

Solve for b

$$b=-4\sqrt{2}\approx -5.656854249$$

$$b=4\sqrt{2}\approx 5.656854249$$

Steps Using the Quadratic Formula

Substitute t for b^{2}.

$$t^{2}-29t-96=0$$

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, -29 for b, and -96 for c in the quadratic formula.

$$t=\frac{-\left(-29\right)±\sqrt{\left(-29\right)^{2}-4\times 1\left(-96\right)}}{2}$$

Do the calculations.

$$t=\frac{29±35}{2}$$

Solve the equation t=\frac{29±35}{2} when ± is plus and when ± is minus.

$$t=32 t=-3$$

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

$$b=4\sqrt{2} b=-4\sqrt{2}$$