Solve 8t^4+10t^2-2=0

$$8 t ^ { 4 } + 10 t ^ { 2 } - 2 = 0$$

Solve for t

$$t=-\frac{\sqrt{2\left(\sqrt{41}-5\right)}}{4}\approx -0.418796525$$

$$t=\frac{\sqrt{2\left(\sqrt{41}-5\right)}}{4}\approx 0.418796525$$

Steps Using the Quadratic Formula

Substitute t for t^{2}.

$$8t^{2}+10t-2=0$$

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 8 for a, 10 for b, and -2 for c in the quadratic formula.

$$t=\frac{-10±\sqrt{10^{2}-4\times 8\left(-2\right)}}{2\times 8}$$

Do the calculations.

$$t=\frac{-10±2\sqrt{41}}{16}$$

Solve the equation t=\frac{-10±2\sqrt{41}}{16} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{41}-5}{8} t=\frac{-\sqrt{41}-5}{8}$$

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

$$t=\frac{\sqrt{\frac{\sqrt{41}-5}{2}}}{2} t=-\frac{\sqrt{\frac{\sqrt{41}-5}{2}}}{2}$$