Solve (t-2)^2+(2t-8)^2+80=(t-6)^2+t^2

$$( t - 2 ) ^ { 2 } + ( 2 t - 8 ) ^ { 2 } + 80 = ( t - 6 ) ^ { 2 } + t ^ { 2 }$$

Solve for t

$$t=\frac{8\sqrt{3}i}{3}+4\approx 4+4.618802154i$$

$$t=-\frac{8\sqrt{3}i}{3}+4\approx 4-4.618802154i$$

Steps Using the Quadratic Formula

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

$$t^{2}-4t+4+\left(2t-8\right)^{2}+80=\left(t-6\right)^{2}+t^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2t-8\right)^{2}.

$$t^{2}-4t+4+4t^{2}-32t+64+80=\left(t-6\right)^{2}+t^{2}$$

Combine t^{2} and 4t^{2} to get 5t^{2}.

$$5t^{2}-4t+4-32t+64+80=\left(t-6\right)^{2}+t^{2}$$

Combine -4t and -32t to get -36t.

$$5t^{2}-36t+4+64+80=\left(t-6\right)^{2}+t^{2}$$

Add 4 and 64 to get 68.

$$5t^{2}-36t+68+80=\left(t-6\right)^{2}+t^{2}$$

Add 68 and 80 to get 148.

$$5t^{2}-36t+148=\left(t-6\right)^{2}+t^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(t-6\right)^{2}.

$$5t^{2}-36t+148=t^{2}-12t+36+t^{2}$$

Combine t^{2} and t^{2} to get 2t^{2}.

$$5t^{2}-36t+148=2t^{2}-12t+36$$

Subtract 2t^{2} from both sides.

$$5t^{2}-36t+148-2t^{2}=-12t+36$$

Combine 5t^{2} and -2t^{2} to get 3t^{2}.

$$3t^{2}-36t+148=-12t+36$$

Add 12t to both sides.

$$3t^{2}-36t+148+12t=36$$

Combine -36t and 12t to get -24t.

$$3t^{2}-24t+148=36$$

Subtract 36 from both sides.

$$3t^{2}-24t+148-36=0$$

Subtract 36 from 148 to get 112.

$$3t^{2}-24t+112=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -24 for b, and 112 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-24\right)±\sqrt{\left(-24\right)^{2}-4\times 3\times 112}}{2\times 3}$$

Square -24.

$$t=\frac{-\left(-24\right)±\sqrt{576-4\times 3\times 112}}{2\times 3}$$

Multiply -4 times 3.

$$t=\frac{-\left(-24\right)±\sqrt{576-12\times 112}}{2\times 3}$$

Multiply -12 times 112.

$$t=\frac{-\left(-24\right)±\sqrt{576-1344}}{2\times 3}$$

Add 576 to -1344.

$$t=\frac{-\left(-24\right)±\sqrt{-768}}{2\times 3}$$

Take the square root of -768.

$$t=\frac{-\left(-24\right)±16\sqrt{3}i}{2\times 3}$$

The opposite of -24 is 24.

$$t=\frac{24±16\sqrt{3}i}{2\times 3}$$

Multiply 2 times 3.

$$t=\frac{24±16\sqrt{3}i}{6}$$

Now solve the equation t=\frac{24±16\sqrt{3}i}{6} when ± is plus. Add 24 to 16i\sqrt{3}.

$$t=\frac{24+16\sqrt{3}i}{6}$$

Divide 24+16i\sqrt{3} by 6.

$$t=\frac{8\sqrt{3}i}{3}+4$$

Now solve the equation t=\frac{24±16\sqrt{3}i}{6} when ± is minus. Subtract 16i\sqrt{3} from 24.

$$t=\frac{-16\sqrt{3}i+24}{6}$$

Divide 24-16i\sqrt{3} by 6.

$$t=-\frac{8\sqrt{3}i}{3}+4$$

The equation is now solved.

$$t=\frac{8\sqrt{3}i}{3}+4 t=-\frac{8\sqrt{3}i}{3}+4$$

Steps for Completing the Square

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

$$t^{2}-4t+4+\left(2t-8\right)^{2}+80=\left(t-6\right)^{2}+t^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2t-8\right)^{2}.

$$t^{2}-4t+4+4t^{2}-32t+64+80=\left(t-6\right)^{2}+t^{2}$$

Combine t^{2} and 4t^{2} to get 5t^{2}.

$$5t^{2}-4t+4-32t+64+80=\left(t-6\right)^{2}+t^{2}$$

Combine -4t and -32t to get -36t.

$$5t^{2}-36t+4+64+80=\left(t-6\right)^{2}+t^{2}$$

Add 4 and 64 to get 68.

$$5t^{2}-36t+68+80=\left(t-6\right)^{2}+t^{2}$$

Add 68 and 80 to get 148.

$$5t^{2}-36t+148=\left(t-6\right)^{2}+t^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(t-6\right)^{2}.

$$5t^{2}-36t+148=t^{2}-12t+36+t^{2}$$

Combine t^{2} and t^{2} to get 2t^{2}.

$$5t^{2}-36t+148=2t^{2}-12t+36$$

Subtract 2t^{2} from both sides.

$$5t^{2}-36t+148-2t^{2}=-12t+36$$

Combine 5t^{2} and -2t^{2} to get 3t^{2}.

$$3t^{2}-36t+148=-12t+36$$

Add 12t to both sides.

$$3t^{2}-36t+148+12t=36$$

Combine -36t and 12t to get -24t.

$$3t^{2}-24t+148=36$$

Subtract 148 from both sides.

$$3t^{2}-24t=36-148$$

Subtract 148 from 36 to get -112.

$$3t^{2}-24t=-112$$

Divide both sides by 3.

$$\frac{3t^{2}-24t}{3}=-\frac{112}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$t^{2}+\left(-\frac{24}{3}\right)t=-\frac{112}{3}$$

Divide -24 by 3.

$$t^{2}-8t=-\frac{112}{3}$$

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-8t+\left(-4\right)^{2}=-\frac{112}{3}+\left(-4\right)^{2}$$

Square -4.

$$t^{2}-8t+16=-\frac{112}{3}+16$$

Add -\frac{112}{3} to 16.

$$t^{2}-8t+16=-\frac{64}{3}$$

Factor t^{2}-8t+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-4\right)^{2}=-\frac{64}{3}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-4\right)^{2}}=\sqrt{-\frac{64}{3}}$$

Simplify.

$$t-4=\frac{8\sqrt{3}i}{3} t-4=-\frac{8\sqrt{3}i}{3}$$

Add 4 to both sides of the equation.

$$t=\frac{8\sqrt{3}i}{3}+4 t=-\frac{8\sqrt{3}i}{3}+4$$