Solve 9y^4-29y^2+20=80

$$9 y ^ { 4 } - 29 y ^ { 2 } + 20 = 80$$

Solve for y (complex solution)

$$y=\frac{i\sqrt{2\left(\sqrt{3001}-29\right)}}{6}\approx 1.196786988i$$

$$y=-\frac{i\sqrt{2\left(\sqrt{3001}-29\right)}}{6}\approx -0-1.196786988i$$

$$y = -\frac{\sqrt{2 {(\sqrt{3001} + 29)}}}{6} \approx -2.157433966$$

$$y = \frac{\sqrt{2 {(\sqrt{3001} + 29)}}}{6} \approx 2.157433966$$

Steps Using the Quadratic Formula

Subtract 80 from both sides.

$$9y^{4}-29y^{2}+20-80=0$$

Subtract 80 from 20 to get -60.

$$9y^{4}-29y^{2}-60=0$$

Substitute t for y^{2}.

$$9t^{2}-29t-60=0$$

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 9 for a, -29 for b, and -60 for c in the quadratic formula.

$$t=\frac{-\left(-29\right)±\sqrt{\left(-29\right)^{2}-4\times 9\left(-60\right)}}{2\times 9}$$

Do the calculations.

$$t=\frac{29±\sqrt{3001}}{18}$$

Solve the equation t=\frac{29±\sqrt{3001}}{18} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{3001}+29}{18} t=\frac{29-\sqrt{3001}}{18}$$

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

$$y=-\sqrt{\frac{\sqrt{3001}+29}{18}} y=\sqrt{\frac{\sqrt{3001}+29}{18}} y=-i\sqrt{-\frac{29-\sqrt{3001}}{18}} y=i\sqrt{-\frac{29-\sqrt{3001}}{18}}$$

Solve for y

$$y = -\frac{\sqrt{2 {(\sqrt{3001} + 29)}}}{6} \approx -2.157433966$$

$$y = \frac{\sqrt{2 {(\sqrt{3001} + 29)}}}{6} \approx 2.157433966$$

Steps Using the Quadratic Formula

Subtract 80 from both sides.

$$9y^{4}-29y^{2}+20-80=0$$

Subtract 80 from 20 to get -60.

$$9y^{4}-29y^{2}-60=0$$

Substitute t for y^{2}.

$$9t^{2}-29t-60=0$$

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 9 for a, -29 for b, and -60 for c in the quadratic formula.

$$t=\frac{-\left(-29\right)±\sqrt{\left(-29\right)^{2}-4\times 9\left(-60\right)}}{2\times 9}$$

Do the calculations.

$$t=\frac{29±\sqrt{3001}}{18}$$

Solve the equation t=\frac{29±\sqrt{3001}}{18} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{3001}+29}{18} t=\frac{29-\sqrt{3001}}{18}$$

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

$$y=\frac{\sqrt{2\sqrt{3001}+58}}{6} y=-\frac{\sqrt{2\sqrt{3001}+58}}{6}$$