Solve (t+2)^2-(2-t)^2=t^2

$$( t + 2 ) ^ { 2 } - ( 2 - t ) ^ { 2 } = t ^ { 2 }$$

Solve for t

$$t=8$$

$$t=0$$

Steps Using Factoring

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

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

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

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

To find the opposite of 4-4t+t^{2}, find the opposite of each term.

$$t^{2}+4t+4-4+4t-t^{2}=t^{2}$$

Subtract 4 from 4 to get 0.

$$t^{2}+4t+4t-t^{2}=t^{2}$$

Combine 4t and 4t to get 8t.

$$t^{2}+8t-t^{2}=t^{2}$$

Combine t^{2} and -t^{2} to get 0.

$$8t=t^{2}$$

Subtract t^{2} from both sides.

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

Factor out t.

$$t\left(8-t\right)=0$$

To find equation solutions, solve t=0 and 8-t=0.

$$t=0 t=8$$

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}.

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

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

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

To find the opposite of 4-4t+t^{2}, find the opposite of each term.

$$t^{2}+4t+4-4+4t-t^{2}=t^{2}$$

Subtract 4 from 4 to get 0.

$$t^{2}+4t+4t-t^{2}=t^{2}$$

Combine 4t and 4t to get 8t.

$$t^{2}+8t-t^{2}=t^{2}$$

Combine t^{2} and -t^{2} to get 0.

$$8t=t^{2}$$

Subtract t^{2} from both sides.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 8 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Take the square root of 8^{2}.

$$t=\frac{-8±8}{2\left(-1\right)}$$

Multiply 2 times -1.

$$t=\frac{-8±8}{-2}$$

Now solve the equation t=\frac{-8±8}{-2} when ± is plus. Add -8 to 8.

$$t=\frac{0}{-2}$$

Divide 0 by -2.

$$t=0$$

Now solve the equation t=\frac{-8±8}{-2} when ± is minus. Subtract 8 from -8.

$$t=-\frac{16}{-2}$$

Divide -16 by -2.

$$t=8$$

The equation is now solved.

$$t=0 t=8$$

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}.

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

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

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

To find the opposite of 4-4t+t^{2}, find the opposite of each term.

$$t^{2}+4t+4-4+4t-t^{2}=t^{2}$$

Subtract 4 from 4 to get 0.

$$t^{2}+4t+4t-t^{2}=t^{2}$$

Combine 4t and 4t to get 8t.

$$t^{2}+8t-t^{2}=t^{2}$$

Combine t^{2} and -t^{2} to get 0.

$$8t=t^{2}$$

Subtract t^{2} from both sides.

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

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

Divide both sides by -1.

$$\frac{-t^{2}+8t}{-1}=\frac{0}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$t^{2}+\frac{8}{-1}t=\frac{0}{-1}$$

Divide 8 by -1.

$$t^{2}-8t=\frac{0}{-1}$$

Divide 0 by -1.

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

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.

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

Square -4.

$$t^{2}-8t+16=16$$

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}.

$$\left(t-4\right)^{2}=16$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-4\right)^{2}}=\sqrt{16}$$

Simplify.

$$t-4=4 t-4=-4$$

Add 4 to both sides of the equation.

$$t=8 t=0$$