Solve t^4+t^2=20

$$t ^ { 4 } + t ^ { 2 } = 20$$

Solve for t

$$t=-2$$

$$t=2$$

Steps Using the Quadratic Formula

Subtract 20 from both sides.

$$t^{4}+t^{2}-20=0$$

Substitute t for t^{2}.

$$t^{2}+t-20=0$$

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, 1 for b, and -20 for c in the quadratic formula.

$$t=\frac{-1±\sqrt{1^{2}-4\times 1\left(-20\right)}}{2}$$

Do the calculations.

$$t=\frac{-1±9}{2}$$

Solve the equation t=\frac{-1±9}{2} when ± is plus and when ± is minus.

$$t=4 t=-5$$

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

$$t=2 t=-2$$