Solve 4t^2+25t^4=32/25

$$4 { t }^{ 2 } +25 { t }^{ 4 } = \frac{ 32 }{ 25 }$$

Solve for t (complex solution)

$$t=\frac{2}{5}=0.4$$

$$t=-\frac{2}{5}=-0.4$$

$$t=-\frac{2\sqrt{2}i}{5}\approx -0-0.565685425i$$

$$t=\frac{2\sqrt{2}i}{5}\approx 0.565685425i$$

Steps Using the Quadratic Formula

Subtract \frac{32}{25} from both sides.

$$4t^{2}+25t^{4}-\frac{32}{25}=0$$

Substitute t for t^{2}.

$$25t^{2}+4t-\frac{32}{25}=0$$

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 25 for a, 4 for b, and -\frac{32}{25} for c in the quadratic formula.

$$t=\frac{-4±\sqrt{4^{2}-4\times 25\left(-\frac{32}{25}\right)}}{2\times 25}$$

Do the calculations.

$$t=\frac{-4±12}{50}$$

Solve the equation t=\frac{-4±12}{50} when ± is plus and when ± is minus.

$$t=\frac{4}{25} t=-\frac{8}{25}$$

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

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

Solve for t

$$t=-\frac{2}{5}=-0.4$$

$$t=\frac{2}{5}=0.4$$

Steps Using the Quadratic Formula

Subtract \frac{32}{25} from both sides.

$$4t^{2}+25t^{4}-\frac{32}{25}=0$$

Substitute t for t^{2}.

$$25t^{2}+4t-\frac{32}{25}=0$$

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 25 for a, 4 for b, and -\frac{32}{25} for c in the quadratic formula.

$$t=\frac{-4±\sqrt{4^{2}-4\times 25\left(-\frac{32}{25}\right)}}{2\times 25}$$

Do the calculations.

$$t=\frac{-4±12}{50}$$

Solve the equation t=\frac{-4±12}{50} when ± is plus and when ± is minus.

$$t=\frac{4}{25} t=-\frac{8}{25}$$

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

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