Solve 3y^4-5y^2+1.3=0

$$3 y ^ { 4 } - 5 y ^ { 2 } + 1.3 = 0$$

Solve for y

$$y=-\frac{\sqrt{750-30\sqrt{235}}}{30}\approx -0.567752595$$

$$y=\frac{\sqrt{750-30\sqrt{235}}}{30}\approx 0.567752595$$

$$y = \frac{\sqrt{30 \sqrt{235} + 750}}{30} \approx 1.159449722$$

$$y = -\frac{\sqrt{30 \sqrt{235} + 750}}{30} \approx -1.159449722$$

Steps Using the Quadratic Formula

Substitute t for y^{2}.

$$3t^{2}-5t+1.3=0$$

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 3 for a, -5 for b, and 1.3 for c in the quadratic formula.

$$t=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 3\times 1.3}}{2\times 3}$$

Do the calculations.

$$t=\frac{5±\frac{1}{5}\sqrt{235}}{6}$$

Solve the equation t=\frac{5±\frac{1}{5}\sqrt{235}}{6} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{235}}{30}+\frac{5}{6} t=-\frac{\sqrt{235}}{30}+\frac{5}{6}$$

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

$$y=\frac{\sqrt{\frac{2\sqrt{235}}{15}+\frac{10}{3}}}{2} y=-\frac{\sqrt{\frac{2\sqrt{235}}{15}+\frac{10}{3}}}{2} y=\frac{\sqrt{-\frac{2\sqrt{235}}{15}+\frac{10}{3}}}{2} y=-\frac{\sqrt{-\frac{2\sqrt{235}}{15}+\frac{10}{3}}}{2}$$