Solve 2t^2=(t+3)^2

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

Solve for t

$$t=3\sqrt{2}+3\approx 7.242640687$$

$$t=3-3\sqrt{2}\approx -1.242640687$$

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}.

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

Subtract t^{2} from both sides.

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

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

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

Subtract 6t from both sides.

$$t^{2}-6t=9$$

Subtract 9 from both sides.

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

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}.

$$t=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\left(-9\right)}}{2}$$

Square -6.

$$t=\frac{-\left(-6\right)±\sqrt{36-4\left(-9\right)}}{2}$$

Multiply -4 times -9.

$$t=\frac{-\left(-6\right)±\sqrt{36+36}}{2}$$

Add 36 to 36.

$$t=\frac{-\left(-6\right)±\sqrt{72}}{2}$$

Take the square root of 72.

$$t=\frac{-\left(-6\right)±6\sqrt{2}}{2}$$

The opposite of -6 is 6.

$$t=\frac{6±6\sqrt{2}}{2}$$

Now solve the equation t=\frac{6±6\sqrt{2}}{2} when ± is plus. Add 6 to 6\sqrt{2}.

$$t=\frac{6\sqrt{2}+6}{2}$$

Divide 6+6\sqrt{2} by 2.

$$t=3\sqrt{2}+3$$

Now solve the equation t=\frac{6±6\sqrt{2}}{2} when ± is minus. Subtract 6\sqrt{2} from 6.

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

Divide 6-6\sqrt{2} by 2.

$$t=3-3\sqrt{2}$$

The equation is now solved.

$$t=3\sqrt{2}+3 t=3-3\sqrt{2}$$

Steps for Completing the Square

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

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

Subtract t^{2} from both sides.

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

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

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

Subtract 6t from both sides.

$$t^{2}-6t=9$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-6t+\left(-3\right)^{2}=9+\left(-3\right)^{2}$$

Square -3.

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

Add 9 to 9.

$$t^{2}-6t+9=18$$

Factor t^{2}-6t+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-3\right)^{2}=18$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-3\right)^{2}}=\sqrt{18}$$

Simplify.

$$t-3=3\sqrt{2} t-3=-3\sqrt{2}$$

Add 3 to both sides of the equation.

$$t=3\sqrt{2}+3 t=3-3\sqrt{2}$$