Solve 2y^2-8/14y^2=6-2y^2

$$\frac { 2 y ^ { 2 } - 8 } { 14 y ^ { 2 } } = 6 - 2 y ^ { 2 }$$

Solve for y (complex solution)

$$y = \frac{\sqrt{7 {(\sqrt{1905} + 41)}}}{14} \approx 1.738701339$$

$$y = -\frac{\sqrt{7 {(\sqrt{1905} + 41)}}}{14} \approx -1.738701339$$

$$y=-\frac{i\sqrt{7\left(\sqrt{1905}-41\right)}}{14}\approx -0-0.30742628i$$

$$y=\frac{i\sqrt{7\left(\sqrt{1905}-41\right)}}{14}\approx 0.30742628i$$

Steps Using the Quadratic Formula

Variable y cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 14y^{2}.

$$2y^{2}-8=14y^{2}\times 6-2y^{2}\times 14y^{2}$$

To multiply powers of the same base, add their exponents. Add 2 and 2 to get 4.

$$2y^{2}-8=14y^{2}\times 6-2y^{4}\times 14$$

Multiply 14 and 6 to get 84.

$$2y^{2}-8=84y^{2}-2y^{4}\times 14$$

Multiply -2 and 14 to get -28.

$$2y^{2}-8=84y^{2}-28y^{4}$$

Subtract 84y^{2} from both sides.

$$2y^{2}-8-84y^{2}=-28y^{4}$$

Combine 2y^{2} and -84y^{2} to get -82y^{2}.

$$-82y^{2}-8=-28y^{4}$$

Add 28y^{4} to both sides.

$$-82y^{2}-8+28y^{4}=0$$

Substitute t for y^{2}.

$$28t^{2}-82t-8=0$$

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 28 for a, -82 for b, and -8 for c in the quadratic formula.

$$t=\frac{-\left(-82\right)±\sqrt{\left(-82\right)^{2}-4\times 28\left(-8\right)}}{2\times 28}$$

Do the calculations.

$$t=\frac{82±2\sqrt{1905}}{56}$$

Solve the equation t=\frac{82±2\sqrt{1905}}{56} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{1905}+41}{28} t=\frac{41-\sqrt{1905}}{28}$$

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

$$y=-\sqrt{\frac{\sqrt{1905}+41}{28}} y=\sqrt{\frac{\sqrt{1905}+41}{28}} y=-i\sqrt{-\frac{41-\sqrt{1905}}{28}} y=i\sqrt{-\frac{41-\sqrt{1905}}{28}}$$

Solve for y

$$y = \frac{\sqrt{7 {(\sqrt{1905} + 41)}}}{14} \approx 1.738701339$$

$$y = -\frac{\sqrt{7 {(\sqrt{1905} + 41)}}}{14} \approx -1.738701339$$

Steps Using the Quadratic Formula

Variable y cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 14y^{2}.

$$2y^{2}-8=14y^{2}\times 6-2y^{2}\times 14y^{2}$$

To multiply powers of the same base, add their exponents. Add 2 and 2 to get 4.

$$2y^{2}-8=14y^{2}\times 6-2y^{4}\times 14$$

Multiply 14 and 6 to get 84.

$$2y^{2}-8=84y^{2}-2y^{4}\times 14$$

Multiply -2 and 14 to get -28.

$$2y^{2}-8=84y^{2}-28y^{4}$$

Subtract 84y^{2} from both sides.

$$2y^{2}-8-84y^{2}=-28y^{4}$$

Combine 2y^{2} and -84y^{2} to get -82y^{2}.

$$-82y^{2}-8=-28y^{4}$$

Add 28y^{4} to both sides.

$$-82y^{2}-8+28y^{4}=0$$

Substitute t for y^{2}.

$$28t^{2}-82t-8=0$$

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 28 for a, -82 for b, and -8 for c in the quadratic formula.

$$t=\frac{-\left(-82\right)±\sqrt{\left(-82\right)^{2}-4\times 28\left(-8\right)}}{2\times 28}$$

Do the calculations.

$$t=\frac{82±2\sqrt{1905}}{56}$$

Solve the equation t=\frac{82±2\sqrt{1905}}{56} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{1905}+41}{28} t=\frac{41-\sqrt{1905}}{28}$$

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

$$y=\frac{\sqrt{\frac{\sqrt{1905}+41}{7}}}{2} y=-\frac{\sqrt{\frac{\sqrt{1905}+41}{7}}}{2}$$