Solve 10=28t^2-6t^4

$$10 = 28 t ^ { 2 } - 6 t ^ { 4 }$$

Solve for t

$$t=-\frac{\sqrt{21-3\sqrt{34}}}{3}\approx -0.624245706$$

$$t=\frac{\sqrt{21-3\sqrt{34}}}{3}\approx 0.624245706$$

$$t = \frac{\sqrt{3 \sqrt{34} + 21}}{3} \approx 2.06808703$$

$$t = -\frac{\sqrt{3 \sqrt{34} + 21}}{3} \approx -2.06808703$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$28t^{2}-6t^{4}=10$$

Subtract 10 from both sides.

$$28t^{2}-6t^{4}-10=0$$

Substitute t for t^{2}.

$$-6t^{2}+28t-10=0$$

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 -6 for a, 28 for b, and -10 for c in the quadratic formula.

$$t=\frac{-28±\sqrt{28^{2}-4\left(-6\right)\left(-10\right)}}{-6\times 2}$$

Do the calculations.

$$t=\frac{-28±4\sqrt{34}}{-12}$$

Solve the equation t=\frac{-28±4\sqrt{34}}{-12} when ± is plus and when ± is minus.

$$t=\frac{7-\sqrt{34}}{3} t=\frac{\sqrt{34}+7}{3}$$

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

$$t=\sqrt{\frac{7-\sqrt{34}}{3}} t=-\sqrt{\frac{7-\sqrt{34}}{3}} t=\sqrt{\frac{\sqrt{34}+7}{3}} t=-\sqrt{\frac{\sqrt{34}+7}{3}}$$