Solve 9(1-t)^2=12t^2+(t-3)^2

$$9 ( 1 - t ) ^ { 2 } = 12 t ^ { 2 } + ( t - 3 ) ^ { 2 }$$

Solve for t

$$t=-3$$

$$t=0$$

Steps Using Factoring

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-t\right)^{2}.

$$9\left(1-2t+t^{2}\right)=12t^{2}+\left(t-3\right)^{2}$$

Use the distributive property to multiply 9 by 1-2t+t^{2}.

$$9-18t+9t^{2}=12t^{2}+\left(t-3\right)^{2}$$

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

$$9-18t+9t^{2}=12t^{2}+t^{2}-6t+9$$

Combine 12t^{2} and t^{2} to get 13t^{2}.

$$9-18t+9t^{2}=13t^{2}-6t+9$$

Subtract 13t^{2} from both sides.

$$9-18t+9t^{2}-13t^{2}=-6t+9$$

Combine 9t^{2} and -13t^{2} to get -4t^{2}.

$$9-18t-4t^{2}=-6t+9$$

Add 6t to both sides.

$$9-18t-4t^{2}+6t=9$$

Combine -18t and 6t to get -12t.

$$9-12t-4t^{2}=9$$

Subtract 9 from both sides.

$$9-12t-4t^{2}-9=0$$

Subtract 9 from 9 to get 0.

$$-12t-4t^{2}=0$$

Factor out t.

$$t\left(-12-4t\right)=0$$

To find equation solutions, solve t=0 and -12-4t=0.

$$t=0 t=-3$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-t\right)^{2}.

$$9\left(1-2t+t^{2}\right)=12t^{2}+\left(t-3\right)^{2}$$

Use the distributive property to multiply 9 by 1-2t+t^{2}.

$$9-18t+9t^{2}=12t^{2}+\left(t-3\right)^{2}$$

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

$$9-18t+9t^{2}=12t^{2}+t^{2}-6t+9$$

Combine 12t^{2} and t^{2} to get 13t^{2}.

$$9-18t+9t^{2}=13t^{2}-6t+9$$

Subtract 13t^{2} from both sides.

$$9-18t+9t^{2}-13t^{2}=-6t+9$$

Combine 9t^{2} and -13t^{2} to get -4t^{2}.

$$9-18t-4t^{2}=-6t+9$$

Add 6t to both sides.

$$9-18t-4t^{2}+6t=9$$

Combine -18t and 6t to get -12t.

$$9-12t-4t^{2}=9$$

Subtract 9 from both sides.

$$9-12t-4t^{2}-9=0$$

Subtract 9 from 9 to get 0.

$$-12t-4t^{2}=0$$

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.

$$-4t^{2}-12t=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, -12 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}}}{2\left(-4\right)}$$

Take the square root of \left(-12\right)^{2}.

$$t=\frac{-\left(-12\right)±12}{2\left(-4\right)}$$

The opposite of -12 is 12.

$$t=\frac{12±12}{2\left(-4\right)}$$

Multiply 2 times -4.

$$t=\frac{12±12}{-8}$$

Now solve the equation t=\frac{12±12}{-8} when ± is plus. Add 12 to 12.

$$t=\frac{24}{-8}$$

Divide 24 by -8.

$$t=-3$$

Now solve the equation t=\frac{12±12}{-8} when ± is minus. Subtract 12 from 12.

$$t=\frac{0}{-8}$$

Divide 0 by -8.

$$t=0$$

The equation is now solved.

$$t=-3 t=0$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-t\right)^{2}.

$$9\left(1-2t+t^{2}\right)=12t^{2}+\left(t-3\right)^{2}$$

Use the distributive property to multiply 9 by 1-2t+t^{2}.

$$9-18t+9t^{2}=12t^{2}+\left(t-3\right)^{2}$$

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

$$9-18t+9t^{2}=12t^{2}+t^{2}-6t+9$$

Combine 12t^{2} and t^{2} to get 13t^{2}.

$$9-18t+9t^{2}=13t^{2}-6t+9$$

Subtract 13t^{2} from both sides.

$$9-18t+9t^{2}-13t^{2}=-6t+9$$

Combine 9t^{2} and -13t^{2} to get -4t^{2}.

$$9-18t-4t^{2}=-6t+9$$

Add 6t to both sides.

$$9-18t-4t^{2}+6t=9$$

Combine -18t and 6t to get -12t.

$$9-12t-4t^{2}=9$$

Subtract 9 from both sides.

$$-12t-4t^{2}=9-9$$

Subtract 9 from 9 to get 0.

$$-12t-4t^{2}=0$$

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.

$$-4t^{2}-12t=0$$

Divide both sides by -4.

$$\frac{-4t^{2}-12t}{-4}=\frac{0}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$t^{2}+\left(-\frac{12}{-4}\right)t=\frac{0}{-4}$$

Divide -12 by -4.

$$t^{2}+3t=\frac{0}{-4}$$

Divide 0 by -4.

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

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

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

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$t^{2}+3t+\frac{9}{4}=\frac{9}{4}$$

Factor t^{2}+3t+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t+\frac{3}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t+\frac{3}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

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

Subtract \frac{3}{2} from both sides of the equation.

$$t=0 t=-3$$