Solve z^2/6-3z/2-6=0

$$\frac { z ^ { 2 } } { 6 } - \frac { 3 z } { 2 } - 6 = 0$$

Solve for z

$$z=-3$$

$$z=12$$

Steps Using Factoring

Multiply both sides of the equation by 6, the least common multiple of 6,2.

$$z^{2}-3\times 3z-36=0$$

Multiply -3 and 3 to get -9.

$$z^{2}-9z-36=0$$

To solve the equation, factor z^{2}-9z-36 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.

$$a+b=-9 ab=-36$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -36.

$$1,-36 2,-18 3,-12 4,-9 6,-6$$

Calculate the sum for each pair.

$$1-36=-35 2-18=-16 3-12=-9 4-9=-5 6-6=0$$

The solution is the pair that gives sum -9.

$$a=-12 b=3$$

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

$$\left(z-12\right)\left(z+3\right)$$

To find equation solutions, solve z-12=0 and z+3=0.

$$z=12 z=-3$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 6, the least common multiple of 6,2.

$$z^{2}-3\times 3z-36=0$$

Multiply -3 and 3 to get -9.

$$z^{2}-9z-36=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as z^{2}+az+bz-36. To find a and b, set up a system to be solved.

$$a+b=-9 ab=1\left(-36\right)=-36$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -36.

$$1,-36 2,-18 3,-12 4,-9 6,-6$$

Calculate the sum for each pair.

$$1-36=-35 2-18=-16 3-12=-9 4-9=-5 6-6=0$$

The solution is the pair that gives sum -9.

$$a=-12 b=3$$

Rewrite z^{2}-9z-36 as \left(z^{2}-12z\right)+\left(3z-36\right).

$$\left(z^{2}-12z\right)+\left(3z-36\right)$$

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

$$z\left(z-12\right)+3\left(z-12\right)$$

Factor out common term z-12 by using distributive property.

$$\left(z-12\right)\left(z+3\right)$$

To find equation solutions, solve z-12=0 and z+3=0.

$$z=12 z=-3$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 6, the least common multiple of 6,2.

$$z^{2}-3\times 3z-36=0$$

Multiply -3 and 3 to get -9.

$$z^{2}-9z-36=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -9 for b, and -36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -9.

$$z=\frac{-\left(-9\right)±\sqrt{81-4\left(-36\right)}}{2}$$

Multiply -4 times -36.

$$z=\frac{-\left(-9\right)±\sqrt{81+144}}{2}$$

Add 81 to 144.

$$z=\frac{-\left(-9\right)±\sqrt{225}}{2}$$

Take the square root of 225.

$$z=\frac{-\left(-9\right)±15}{2}$$

The opposite of -9 is 9.

$$z=\frac{9±15}{2}$$

Now solve the equation z=\frac{9±15}{2} when ± is plus. Add 9 to 15.

$$z=\frac{24}{2}$$

Divide 24 by 2.

$$z=12$$

Now solve the equation z=\frac{9±15}{2} when ± is minus. Subtract 15 from 9.

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

Divide -6 by 2.

$$z=-3$$

The equation is now solved.

$$z=12 z=-3$$

Steps for Completing the Square

Multiply both sides of the equation by 6, the least common multiple of 6,2.

$$z^{2}-3\times 3z-36=0$$

Multiply -3 and 3 to get -9.

$$z^{2}-9z-36=0$$

Add 36 to both sides. Anything plus zero gives itself.

$$z^{2}-9z=36$$

Divide -9, the coefficient of the x term, by 2 to get -\frac{9}{2}. Then add the square of -\frac{9}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$z^{2}-9z+\left(-\frac{9}{2}\right)^{2}=36+\left(-\frac{9}{2}\right)^{2}$$

Square -\frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$z^{2}-9z+\frac{81}{4}=36+\frac{81}{4}$$

Add 36 to \frac{81}{4}.

$$z^{2}-9z+\frac{81}{4}=\frac{225}{4}$$

Factor z^{2}-9z+\frac{81}{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}.

$$\left(z-\frac{9}{2}\right)^{2}=\frac{225}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(z-\frac{9}{2}\right)^{2}}=\sqrt{\frac{225}{4}}$$

Simplify.

$$z-\frac{9}{2}=\frac{15}{2} z-\frac{9}{2}=-\frac{15}{2}$$

Add \frac{9}{2} to both sides of the equation.

$$z=12 z=-3$$