Solve (t+3)^2=0

$$( t + 3 ) ^ { 2 } = 0$$

Solve for t

$$t=-3$$

Steps Using Factoring

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(t+3\right)^{2}.

$$t^{2}+6t+9=0$$

To solve the equation, factor t^{2}+6t+9 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.

$$a+b=6 ab=9$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 9.

$$1,9 3,3$$

Calculate the sum for each pair.

$$1+9=10 3+3=6$$

The solution is the pair that gives sum 6.

$$a=3 b=3$$

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

$$\left(t+3\right)\left(t+3\right)$$

Rewrite as a binomial square.

$$\left(t+3\right)^{2}$$

To find equation solution, solve t+3=0.

$$t=-3$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(t+3\right)^{2}.

$$t^{2}+6t+9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as t^{2}+at+bt+9. To find a and b, set up a system to be solved.

$$a+b=6 ab=1\times 9=9$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 9.

$$1,9 3,3$$

Calculate the sum for each pair.

$$1+9=10 3+3=6$$

The solution is the pair that gives sum 6.

$$a=3 b=3$$

Rewrite t^{2}+6t+9 as \left(t^{2}+3t\right)+\left(3t+9\right).

$$\left(t^{2}+3t\right)+\left(3t+9\right)$$

Factor out t in the first and 3 in the second group.

$$t\left(t+3\right)+3\left(t+3\right)$$

Factor out common term t+3 by using distributive property.

$$\left(t+3\right)\left(t+3\right)$$

Rewrite as a binomial square.

$$\left(t+3\right)^{2}$$

To find equation solution, solve t+3=0.

$$t=-3$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(t+3\right)^{2}.

$$t^{2}+6t+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 6 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-6±\sqrt{6^{2}-4\times 9}}{2}$$

Square 6.

$$t=\frac{-6±\sqrt{36-4\times 9}}{2}$$

Multiply -4 times 9.

$$t=\frac{-6±\sqrt{36-36}}{2}$$

Add 36 to -36.

$$t=\frac{-6±\sqrt{0}}{2}$$

Take the square root of 0.

$$t=-\frac{6}{2}$$

Divide -6 by 2.

$$t=-3$$

Steps for Completing the Square

Take the square root of both sides of the equation.

$$\sqrt{\left(t+3\right)^{2}}=\sqrt{0}$$

Simplify.

$$t+3=0 t+3=0$$

Subtract 3 from both sides of the equation.

$$t=-3 t=-3$$

The equation is now solved. Solutions are the same.

$$t=-3$$