Solve z^2+32=12z

$$z ^ { 2 } + 32 = 12 z$$

Solve for z

$$z=4$$

$$z=8$$

Steps Using Factoring

Subtract 12z from both sides.

$$z^{2}+32-12z=0$$

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

$$z^{2}-12z+32=0$$

To solve the equation, factor z^{2}-12z+32 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.

$$a+b=-12 ab=32$$

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 32.

$$-1,-32 -2,-16 -4,-8$$

Calculate the sum for each pair.

$$-1-32=-33 -2-16=-18 -4-8=-12$$

The solution is the pair that gives sum -12.

$$a=-8 b=-4$$

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

$$\left(z-8\right)\left(z-4\right)$$

To find equation solutions, solve z-8=0 and z-4=0.

$$z=8 z=4$$

Steps Using Factoring By Grouping

Subtract 12z from both sides.

$$z^{2}+32-12z=0$$

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

$$z^{2}-12z+32=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as z^{2}+az+bz+32. To find a and b, set up a system to be solved.

$$a+b=-12 ab=1\times 32=32$$

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 32.

$$-1,-32 -2,-16 -4,-8$$

Calculate the sum for each pair.

$$-1-32=-33 -2-16=-18 -4-8=-12$$

The solution is the pair that gives sum -12.

$$a=-8 b=-4$$

Rewrite z^{2}-12z+32 as \left(z^{2}-8z\right)+\left(-4z+32\right).

$$\left(z^{2}-8z\right)+\left(-4z+32\right)$$

Factor out z in the first and -4 in the second group.

$$z\left(z-8\right)-4\left(z-8\right)$$

Factor out common term z-8 by using distributive property.

$$\left(z-8\right)\left(z-4\right)$$

To find equation solutions, solve z-8=0 and z-4=0.

$$z=8 z=4$$

Steps Using the Quadratic Formula

Subtract 12z from both sides.

$$z^{2}+32-12z=0$$

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.

$$z^{2}-12z+32=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -12 for b, and 32 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$z=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 32}}{2}$$

Square -12.

$$z=\frac{-\left(-12\right)±\sqrt{144-4\times 32}}{2}$$

Multiply -4 times 32.

$$z=\frac{-\left(-12\right)±\sqrt{144-128}}{2}$$

Add 144 to -128.

$$z=\frac{-\left(-12\right)±\sqrt{16}}{2}$$

Take the square root of 16.

$$z=\frac{-\left(-12\right)±4}{2}$$

The opposite of -12 is 12.

$$z=\frac{12±4}{2}$$

Now solve the equation z=\frac{12±4}{2} when ± is plus. Add 12 to 4.

$$z=\frac{16}{2}$$

Divide 16 by 2.

$$z=8$$

Now solve the equation z=\frac{12±4}{2} when ± is minus. Subtract 4 from 12.

$$z=\frac{8}{2}$$

Divide 8 by 2.

$$z=4$$

The equation is now solved.

$$z=8 z=4$$

Steps for Completing the Square

Subtract 12z from both sides.

$$z^{2}+32-12z=0$$

Subtract 32 from both sides. Anything subtracted from zero gives its negation.

$$z^{2}-12z=-32$$

Divide -12, the coefficient of the x term, by 2 to get -6. Then add the square of -6 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$z^{2}-12z+\left(-6\right)^{2}=-32+\left(-6\right)^{2}$$

Square -6.

$$z^{2}-12z+36=-32+36$$

Add -32 to 36.

$$z^{2}-12z+36=4$$

Factor z^{2}-12z+36. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(z-6\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(z-6\right)^{2}}=\sqrt{4}$$

Simplify.

$$z-6=2 z-6=-2$$

Add 6 to both sides of the equation.

$$z=8 z=4$$