Solve -144/25+6/5y^2+y^4=0

$$- \frac { 144 } { 25 } + \frac { 6 } { 5 } y ^ { 2 } + y ^ { 4 } = 0$$

Solve for y (complex solution)

$$y=\frac{i\sqrt{15\left(\sqrt{17}+1\right)}}{5}\approx 1.753243673i$$

$$y=-\frac{i\sqrt{15\left(\sqrt{17}+1\right)}}{5}\approx -0-1.753243673i$$

$$y = -\frac{\sqrt{15 {(\sqrt{17} - 1)}}}{5} \approx -1.368891294$$

$$y = \frac{\sqrt{15 {(\sqrt{17} - 1)}}}{5} \approx 1.368891294$$

Steps Using the Quadratic Formula

Substitute t for y^{2}.

$$t^{2}+\frac{6}{5}t-\frac{144}{25}=0$$

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 1 for a, \frac{6}{5} for b, and -\frac{144}{25} for c in the quadratic formula.

$$t=\frac{-\frac{6}{5}±\sqrt{\left(\frac{6}{5}\right)^{2}-4\times 1\left(-\frac{144}{25}\right)}}{2}$$

Do the calculations.

$$t=\frac{-\frac{6}{5}±\frac{6}{5}\sqrt{17}}{2}$$

Solve the equation t=\frac{-\frac{6}{5}±\frac{6}{5}\sqrt{17}}{2} when ± is plus and when ± is minus.

$$t=\frac{3\sqrt{17}-3}{5} t=\frac{-3\sqrt{17}-3}{5}$$

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

$$y=-\sqrt{\frac{3\sqrt{17}-3}{5}} y=\sqrt{\frac{3\sqrt{17}-3}{5}} y=-i\sqrt{\frac{3\sqrt{17}+3}{5}} y=i\sqrt{\frac{3\sqrt{17}+3}{5}}$$

Solve for y

$$y = -\frac{\sqrt{15 {(\sqrt{17} - 1)}}}{5} \approx -1.368891294$$

$$y = \frac{\sqrt{15 {(\sqrt{17} - 1)}}}{5} \approx 1.368891294$$

Steps Using the Quadratic Formula

Substitute t for y^{2}.

$$t^{2}+\frac{6}{5}t-\frac{144}{25}=0$$

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 1 for a, \frac{6}{5} for b, and -\frac{144}{25} for c in the quadratic formula.

$$t=\frac{-\frac{6}{5}±\sqrt{\left(\frac{6}{5}\right)^{2}-4\times 1\left(-\frac{144}{25}\right)}}{2}$$

Do the calculations.

$$t=\frac{-\frac{6}{5}±\frac{6}{5}\sqrt{17}}{2}$$

Solve the equation t=\frac{-\frac{6}{5}±\frac{6}{5}\sqrt{17}}{2} when ± is plus and when ± is minus.

$$t=\frac{3\sqrt{17}-3}{5} t=\frac{-3\sqrt{17}-3}{5}$$

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

$$y=\sqrt{\frac{3\sqrt{17}-3}{5}} y=-\sqrt{\frac{3\sqrt{17}-3}{5}}$$