Solve (y^2-7)^2+6(y^2-7)-16=0

$$( y ^ { 2 } - 7 ) ^ { 2 } + 6 ( y ^ { 2 } - 7 ) - 16 = 0$$

Solve for y (complex solution)

$$y=3$$

$$y=-3$$

$$y=-i$$

$$y=i$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y^{2}-7\right)^{2}.

$$\left(y^{2}\right)^{2}-14y^{2}+49+6\left(y^{2}-7\right)-16=0$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$y^{4}-14y^{2}+49+6\left(y^{2}-7\right)-16=0$$

Use the distributive property to multiply 6 by y^{2}-7.

$$y^{4}-14y^{2}+49+6y^{2}-42-16=0$$

Combine -14y^{2} and 6y^{2} to get -8y^{2}.

$$y^{4}-8y^{2}+49-42-16=0$$

Subtract 42 from 49 to get 7.

$$y^{4}-8y^{2}+7-16=0$$

Subtract 16 from 7 to get -9.

$$y^{4}-8y^{2}-9=0$$

Substitute t for y^{2}.

$$t^{2}-8t-9=0$$

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, -8 for b, and -9 for c in the quadratic formula.

$$t=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 1\left(-9\right)}}{2}$$

Do the calculations.

$$t=\frac{8±10}{2}$$

Solve the equation t=\frac{8±10}{2} when ± is plus and when ± is minus.

$$t=9 t=-1$$

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

$$y=-3 y=3 y=-i y=i$$

Solve for y

$$y=-3$$

$$y=3$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y^{2}-7\right)^{2}.

$$\left(y^{2}\right)^{2}-14y^{2}+49+6\left(y^{2}-7\right)-16=0$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$y^{4}-14y^{2}+49+6\left(y^{2}-7\right)-16=0$$

Use the distributive property to multiply 6 by y^{2}-7.

$$y^{4}-14y^{2}+49+6y^{2}-42-16=0$$

Combine -14y^{2} and 6y^{2} to get -8y^{2}.

$$y^{4}-8y^{2}+49-42-16=0$$

Subtract 42 from 49 to get 7.

$$y^{4}-8y^{2}+7-16=0$$

Subtract 16 from 7 to get -9.

$$y^{4}-8y^{2}-9=0$$

Substitute t for y^{2}.

$$t^{2}-8t-9=0$$

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, -8 for b, and -9 for c in the quadratic formula.

$$t=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 1\left(-9\right)}}{2}$$

Do the calculations.

$$t=\frac{8±10}{2}$$

Solve the equation t=\frac{8±10}{2} when ± is plus and when ± is minus.

$$t=9 t=-1$$

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

$$y=3 y=-3$$