Solve z^2-3=(3z+1)

$$z ^ { 2 } - 3 = ( 3 z + 1 )$$

Solve for z

$$z=-1$$

$$z=4$$

Steps Using Factoring

Subtract 3z from both sides.

$$z^{2}-3-3z=1$$

Subtract 1 from both sides.

$$z^{2}-3-3z-1=0$$

Subtract 1 from -3 to get -4.

$$z^{2}-4-3z=0$$

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

$$z^{2}-3z-4=0$$

To solve the equation, factor z^{2}-3z-4 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.

$$a+b=-3 ab=-4$$

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 -4.

$$1,-4 2,-2$$

Calculate the sum for each pair.

$$1-4=-3 2-2=0$$

The solution is the pair that gives sum -3.

$$a=-4 b=1$$

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

$$\left(z-4\right)\left(z+1\right)$$

To find equation solutions, solve z-4=0 and z+1=0.

$$z=4 z=-1$$

Steps Using Factoring By Grouping

Subtract 3z from both sides.

$$z^{2}-3-3z=1$$

Subtract 1 from both sides.

$$z^{2}-3-3z-1=0$$

Subtract 1 from -3 to get -4.

$$z^{2}-4-3z=0$$

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

$$z^{2}-3z-4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as z^{2}+az+bz-4. To find a and b, set up a system to be solved.

$$a+b=-3 ab=1\left(-4\right)=-4$$

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 -4.

$$1,-4 2,-2$$

Calculate the sum for each pair.

$$1-4=-3 2-2=0$$

The solution is the pair that gives sum -3.

$$a=-4 b=1$$

Rewrite z^{2}-3z-4 as \left(z^{2}-4z\right)+\left(z-4\right).

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

Factor out z in z^{2}-4z.

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

Factor out common term z-4 by using distributive property.

$$\left(z-4\right)\left(z+1\right)$$

To find equation solutions, solve z-4=0 and z+1=0.

$$z=4 z=-1$$

Steps Using the Quadratic Formula

Subtract 3z from both sides.

$$z^{2}-3-3z=1$$

Subtract 1 from both sides.

$$z^{2}-3-3z-1=0$$

Subtract 1 from -3 to get -4.

$$z^{2}-4-3z=0$$

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.

$$z^{2}-3z-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -3 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$z=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-4\right)}}{2}$$

Square -3.

$$z=\frac{-\left(-3\right)±\sqrt{9-4\left(-4\right)}}{2}$$

Multiply -4 times -4.

$$z=\frac{-\left(-3\right)±\sqrt{9+16}}{2}$$

Add 9 to 16.

$$z=\frac{-\left(-3\right)±\sqrt{25}}{2}$$

Take the square root of 25.

$$z=\frac{-\left(-3\right)±5}{2}$$

The opposite of -3 is 3.

$$z=\frac{3±5}{2}$$

Now solve the equation z=\frac{3±5}{2} when ± is plus. Add 3 to 5.

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

Divide 8 by 2.

$$z=4$$

Now solve the equation z=\frac{3±5}{2} when ± is minus. Subtract 5 from 3.

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

Divide -2 by 2.

$$z=-1$$

The equation is now solved.

$$z=4 z=-1$$

Steps for Completing the Square

Subtract 3z from both sides.

$$z^{2}-3-3z=1$$

Add 3 to both sides.

$$z^{2}-3z=1+3$$

Add 1 and 3 to get 4.

$$z^{2}-3z=4$$

Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$z^{2}-3z+\left(-\frac{3}{2}\right)^{2}=4+\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$z^{2}-3z+\frac{9}{4}=4+\frac{9}{4}$$

Add 4 to \frac{9}{4}.

$$z^{2}-3z+\frac{9}{4}=\frac{25}{4}$$

Factor z^{2}-3z+\frac{9}{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}.

$$\left(z-\frac{3}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(z-\frac{3}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$z-\frac{3}{2}=\frac{5}{2} z-\frac{3}{2}=-\frac{5}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$z=4 z=-1$$