Solve (3-a)(3-a)/2=72

$$\frac { ( 3 - a ) ( 3 - a ) } { 2 } = 72$$

Solve for a

$$a=-9$$

$$a=15$$

Steps Using the Quadratic Formula

Multiply both sides by 2.

$$\left(3-a\right)\left(3-a\right)=72\times 2$$

Multiply 3-a and 3-a to get \left(3-a\right)^{2}.

$$\left(3-a\right)^{2}=72\times 2$$

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

$$9-6a+a^{2}=72\times 2$$

Multiply 72 and 2 to get 144.

$$9-6a+a^{2}=144$$

Subtract 144 from both sides.

$$9-6a+a^{2}-144=0$$

Subtract 144 from 9 to get -135.

$$-135-6a+a^{2}=0$$

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.

$$a^{2}-6a-135=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -6 for b, and -135 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\left(-135\right)}}{2}$$

Square -6.

$$a=\frac{-\left(-6\right)±\sqrt{36-4\left(-135\right)}}{2}$$

Multiply -4 times -135.

$$a=\frac{-\left(-6\right)±\sqrt{36+540}}{2}$$

Add 36 to 540.

$$a=\frac{-\left(-6\right)±\sqrt{576}}{2}$$

Take the square root of 576.

$$a=\frac{-\left(-6\right)±24}{2}$$

The opposite of -6 is 6.

$$a=\frac{6±24}{2}$$

Now solve the equation a=\frac{6±24}{2} when ± is plus. Add 6 to 24.

$$a=\frac{30}{2}$$

Divide 30 by 2.

$$a=15$$

Now solve the equation a=\frac{6±24}{2} when ± is minus. Subtract 24 from 6.

$$a=-\frac{18}{2}$$

Divide -18 by 2.

$$a=-9$$

The equation is now solved.

$$a=15 a=-9$$

Steps for Completing the Square

Multiply both sides by 2.

$$\left(3-a\right)\left(3-a\right)=72\times 2$$

Multiply 3-a and 3-a to get \left(3-a\right)^{2}.

$$\left(3-a\right)^{2}=72\times 2$$

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

$$9-6a+a^{2}=72\times 2$$

Multiply 72 and 2 to get 144.

$$9-6a+a^{2}=144$$

Subtract 9 from both sides.

$$-6a+a^{2}=144-9$$

Subtract 9 from 144 to get 135.

$$-6a+a^{2}=135$$

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.

$$a^{2}-6a=135$$

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.

$$a^{2}-6a+\left(-3\right)^{2}=135+\left(-3\right)^{2}$$

Square -3.

$$a^{2}-6a+9=135+9$$

Add 135 to 9.

$$a^{2}-6a+9=144$$

Factor a^{2}-6a+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}.

$$\left(a-3\right)^{2}=144$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-3\right)^{2}}=\sqrt{144}$$

Simplify.

$$a-3=12 a-3=-12$$

Add 3 to both sides of the equation.

$$a=15 a=-9$$