Completing the square
If you've got a quadratic equation on the form of
$$ax^{2}+c=0$$
Then you can solve the equation by using the square root of
$$x=\pm \sqrt{\frac{-c}{a}}$$
Example
$$3x^{2}-243=0$$
$$3x^{2}=243$$
$$x^{2}=\frac{243}{3}$$
$$x^{2}=81$$
$$x=\pm \sqrt{81}$$
$$x=9\: \: or\: \: x=-9$$
This method can only be used if b = 0. If we instead have an equation on the form of
$$x^{2}+bx=0$$
we can't use the square root initially since we do not have c-value. But we can add a constant d to both sides of the equation to get a new equivalent equation that is a perfect square trinomial. Remember that a perfect square trinomial can be written as
$$x^{2}+bx + d=\left ( x+d \right )^{2}=0$$
This process is called completing the square and the constant d we're adding is
$$d=\left (\frac{b}{2} \right )^{2}$$
Example
$$x^{2}+12x=0$$
We begin by finding the constant d that can be used to complete the square.
$$d=\left (\frac{b}{2} \right )^{2}=\left ( \frac{12}{2} \right )^{2}=6^{2}=36$$
$$x^{2}+12x+d=0+d\Rightarrow$$
$$x^{2}+12x+36=0+36\Rightarrow$$
$$\begin{pmatrix}x+6 \end{pmatrix}^{2}=36$$
$$\sqrt{\begin{pmatrix} x+6 \end{pmatrix}^{2}}=\pm \sqrt{36}$$
$$\begin{matrix} x+6=6\: \: &or\: \: & x+6=-6\\ x=0 & & x=-12 \end{matrix}$$
The completing the square method could of course be used to solve quadratic equations on the form of
$$ax^{2}+bx+c=0$$
In this case you will add a constant d that satisfy the formula
$$d=\left ( \frac{b}{2} \right )^{2}-c$$
Video lesson
Solve the equation by completing the squares
$$x^{2} - 3x - 10=0$$