Solve z^2-30=13z

$$z ^ { 2 } - 30 = 13 z$$

Solve for z

$$z=-2$$

$$z=15$$

Steps Using Factoring

Subtract 13z from both sides.

$$z^{2}-30-13z=0$$

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

$$z^{2}-13z-30=0$$

To solve the equation, factor z^{2}-13z-30 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.

$$a+b=-13 ab=-30$$

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 -30.

$$1,-30 2,-15 3,-10 5,-6$$

Calculate the sum for each pair.

$$1-30=-29 2-15=-13 3-10=-7 5-6=-1$$

The solution is the pair that gives sum -13.

$$a=-15 b=2$$

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

$$\left(z-15\right)\left(z+2\right)$$

To find equation solutions, solve z-15=0 and z+2=0.

$$z=15 z=-2$$

Steps Using Factoring By Grouping

Subtract 13z from both sides.

$$z^{2}-30-13z=0$$

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

$$z^{2}-13z-30=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as z^{2}+az+bz-30. To find a and b, set up a system to be solved.

$$a+b=-13 ab=1\left(-30\right)=-30$$

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 -30.

$$1,-30 2,-15 3,-10 5,-6$$

Calculate the sum for each pair.

$$1-30=-29 2-15=-13 3-10=-7 5-6=-1$$

The solution is the pair that gives sum -13.

$$a=-15 b=2$$

Rewrite z^{2}-13z-30 as \left(z^{2}-15z\right)+\left(2z-30\right).

$$\left(z^{2}-15z\right)+\left(2z-30\right)$$

Factor out z in the first and 2 in the second group.

$$z\left(z-15\right)+2\left(z-15\right)$$

Factor out common term z-15 by using distributive property.

$$\left(z-15\right)\left(z+2\right)$$

To find equation solutions, solve z-15=0 and z+2=0.

$$z=15 z=-2$$

Steps Using the Quadratic Formula

Subtract 13z from both sides.

$$z^{2}-30-13z=0$$

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.

$$z^{2}-13z-30=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -13 for b, and -30 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$z=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\left(-30\right)}}{2}$$

Square -13.

$$z=\frac{-\left(-13\right)±\sqrt{169-4\left(-30\right)}}{2}$$

Multiply -4 times -30.

$$z=\frac{-\left(-13\right)±\sqrt{169+120}}{2}$$

Add 169 to 120.

$$z=\frac{-\left(-13\right)±\sqrt{289}}{2}$$

Take the square root of 289.

$$z=\frac{-\left(-13\right)±17}{2}$$

The opposite of -13 is 13.

$$z=\frac{13±17}{2}$$

Now solve the equation z=\frac{13±17}{2} when ± is plus. Add 13 to 17.

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

Divide 30 by 2.

$$z=15$$

Now solve the equation z=\frac{13±17}{2} when ± is minus. Subtract 17 from 13.

$$z=-\frac{4}{2}$$

Divide -4 by 2.

$$z=-2$$

The equation is now solved.

$$z=15 z=-2$$

Steps for Completing the Square

Subtract 13z from both sides.

$$z^{2}-30-13z=0$$

Add 30 to both sides. Anything plus zero gives itself.

$$z^{2}-13z=30$$

Divide -13, the coefficient of the x term, by 2 to get -\frac{13}{2}. Then add the square of -\frac{13}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$z^{2}-13z+\left(-\frac{13}{2}\right)^{2}=30+\left(-\frac{13}{2}\right)^{2}$$

Square -\frac{13}{2} by squaring both the numerator and the denominator of the fraction.

$$z^{2}-13z+\frac{169}{4}=30+\frac{169}{4}$$

Add 30 to \frac{169}{4}.

$$z^{2}-13z+\frac{169}{4}=\frac{289}{4}$$

Factor z^{2}-13z+\frac{169}{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}.

$$\left(z-\frac{13}{2}\right)^{2}=\frac{289}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(z-\frac{13}{2}\right)^{2}}=\sqrt{\frac{289}{4}}$$

Simplify.

$$z-\frac{13}{2}=\frac{17}{2} z-\frac{13}{2}=-\frac{17}{2}$$

Add \frac{13}{2} to both sides of the equation.

$$z=15 z=-2$$