Solve (t-2)^2=0

$$( t - 2 ) ^ { 2 } = 0$$

Solve for t

$$t=2$$

Steps Using Factoring

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

$$t^{2}-4t+4=0$$

To solve the equation, factor t^{2}-4t+4 using formula t^{2}+\left(a+b\right)t+ab=\left(t+a\right)\left(t+b\right). To find a and b, set up a system to be solved.

$$a+b=-4 ab=4$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 4.

$$-1,-4 -2,-2$$

Calculate the sum for each pair.

$$-1-4=-5 -2-2=-4$$

The solution is the pair that gives sum -4.

$$a=-2 b=-2$$

Rewrite factored expression \left(t+a\right)\left(t+b\right) using the obtained values.

$$\left(t-2\right)\left(t-2\right)$$

Rewrite as a binomial square.

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

To find equation solution, solve t-2=0.

$$t=2$$

Steps Using Factoring By Grouping

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

$$t^{2}-4t+4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as t^{2}+at+bt+4. To find a and b, set up a system to be solved.

$$a+b=-4 ab=1\times 4=4$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 4.

$$-1,-4 -2,-2$$

Calculate the sum for each pair.

$$-1-4=-5 -2-2=-4$$

The solution is the pair that gives sum -4.

$$a=-2 b=-2$$

Rewrite t^{2}-4t+4 as \left(t^{2}-2t\right)+\left(-2t+4\right).

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

Factor out t in the first and -2 in the second group.

$$t\left(t-2\right)-2\left(t-2\right)$$

Factor out common term t-2 by using distributive property.

$$\left(t-2\right)\left(t-2\right)$$

Rewrite as a binomial square.

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

To find equation solution, solve t-2=0.

$$t=2$$

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}.

$$t^{2}-4t+4=0$$

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

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

Square -4.

$$t=\frac{-\left(-4\right)±\sqrt{16-4\times 4}}{2}$$

Multiply -4 times 4.

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

Add 16 to -16.

$$t=\frac{-\left(-4\right)±\sqrt{0}}{2}$$

Take the square root of 0.

$$t=-\frac{-4}{2}$$

The opposite of -4 is 4.

$$t=\frac{4}{2}$$

Divide 4 by 2.

$$t=2$$

Steps for Completing the Square

Take the square root of both sides of the equation.

$$\sqrt{\left(t-2\right)^{2}}=\sqrt{0}$$

Simplify.

$$t-2=0 t-2=0$$

Add 2 to both sides of the equation.

$$t=2 t=2$$

The equation is now solved. Solutions are the same.

$$t=2$$