Solve 3(5-z)^2=12

$$3 ( 5 - z ) ^ { 2 } = 12$$

Solve for z

$$z=3$$

$$z=7$$

Steps Using Factoring

Divide both sides by 3.

$$\left(5-z\right)^{2}=\frac{12}{3}$$

Divide 12 by 3 to get 4.

$$\left(5-z\right)^{2}=4$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5-z\right)^{2}.

$$25-10z+z^{2}=4$$

Subtract 4 from both sides.

$$25-10z+z^{2}-4=0$$

Subtract 4 from 25 to get 21.

$$21-10z+z^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$z^{2}-10z+21=0$$

To solve the equation, factor z^{2}-10z+21 using formula z^{2}+\left(a+b\right)z+ab=\left(z+a\right)\left(z+b\right). To find a and b, set up a system to be solved.

$$a+b=-10 ab=21$$

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 21.

$$-1,-21 -3,-7$$

Calculate the sum for each pair.

$$-1-21=-22 -3-7=-10$$

The solution is the pair that gives sum -10.

$$a=-7 b=-3$$

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

$$\left(z-7\right)\left(z-3\right)$$

To find equation solutions, solve z-7=0 and z-3=0.

$$z=7 z=3$$

Steps Using Factoring By Grouping

Divide both sides by 3.

$$\left(5-z\right)^{2}=\frac{12}{3}$$

Divide 12 by 3 to get 4.

$$\left(5-z\right)^{2}=4$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5-z\right)^{2}.

$$25-10z+z^{2}=4$$

Subtract 4 from both sides.

$$25-10z+z^{2}-4=0$$

Subtract 4 from 25 to get 21.

$$21-10z+z^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$z^{2}-10z+21=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as z^{2}+az+bz+21. To find a and b, set up a system to be solved.

$$a+b=-10 ab=1\times 21=21$$

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 21.

$$-1,-21 -3,-7$$

Calculate the sum for each pair.

$$-1-21=-22 -3-7=-10$$

The solution is the pair that gives sum -10.

$$a=-7 b=-3$$

Rewrite z^{2}-10z+21 as \left(z^{2}-7z\right)+\left(-3z+21\right).

$$\left(z^{2}-7z\right)+\left(-3z+21\right)$$

Factor out z in the first and -3 in the second group.

$$z\left(z-7\right)-3\left(z-7\right)$$

Factor out common term z-7 by using distributive property.

$$\left(z-7\right)\left(z-3\right)$$

To find equation solutions, solve z-7=0 and z-3=0.

$$z=7 z=3$$

Steps Using the Quadratic Formula

Divide both sides by 3.

$$\left(5-z\right)^{2}=\frac{12}{3}$$

Divide 12 by 3 to get 4.

$$\left(5-z\right)^{2}=4$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5-z\right)^{2}.

$$25-10z+z^{2}=4$$

Subtract 4 from both sides.

$$25-10z+z^{2}-4=0$$

Subtract 4 from 25 to get 21.

$$21-10z+z^{2}=0$$

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.

$$z^{2}-10z+21=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10 for b, and 21 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$z=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 21}}{2}$$

Square -10.

$$z=\frac{-\left(-10\right)±\sqrt{100-4\times 21}}{2}$$

Multiply -4 times 21.

$$z=\frac{-\left(-10\right)±\sqrt{100-84}}{2}$$

Add 100 to -84.

$$z=\frac{-\left(-10\right)±\sqrt{16}}{2}$$

Take the square root of 16.

$$z=\frac{-\left(-10\right)±4}{2}$$

The opposite of -10 is 10.

$$z=\frac{10±4}{2}$$

Now solve the equation z=\frac{10±4}{2} when ± is plus. Add 10 to 4.

$$z=\frac{14}{2}$$

Divide 14 by 2.

$$z=7$$

Now solve the equation z=\frac{10±4}{2} when ± is minus. Subtract 4 from 10.

$$z=\frac{6}{2}$$

Divide 6 by 2.

$$z=3$$

The equation is now solved.

$$z=7 z=3$$

Steps for Completing the Square

Divide both sides by 3.

$$\left(5-z\right)^{2}=\frac{12}{3}$$

Divide 12 by 3 to get 4.

$$\left(5-z\right)^{2}=4$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5-z\right)^{2}.

$$25-10z+z^{2}=4$$

Subtract 25 from both sides.

$$-10z+z^{2}=4-25$$

Subtract 25 from 4 to get -21.

$$-10z+z^{2}=-21$$

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.

$$z^{2}-10z=-21$$

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$z^{2}-10z+\left(-5\right)^{2}=-21+\left(-5\right)^{2}$$

Square -5.

$$z^{2}-10z+25=-21+25$$

Add -21 to 25.

$$z^{2}-10z+25=4$$

Factor z^{2}-10z+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(z-5\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(z-5\right)^{2}}=\sqrt{4}$$

Simplify.

$$z-5=2 z-5=-2$$

Add 5 to both sides of the equation.

$$z=7 z=3$$