Completing the SquareClick here to go to the quadratic solver page; unlimited worked examples 

Resources 
What is Completing the Square?Completing the square (CTS) is a method for solving quadratic equations in the form \[x^2+bx+c=0\] Something to work up from:
First, we need to look at what happens when we expand a squared bracket with two terms:
\[(x+d)^2 = (x+d)(x+d) = x^2+2dx+d^2\]
The \(2d\) here is the most important part, we will see why shortly.
Explaining with example:
Obviously it isn't usually that easy, if it was then we'd just factorise and solve!
Let's just see where this path takes us for a few more steps, using the example \[x^2+8x+12=0\]
We think about factorising this into a single bracket, we know that it would be something like
\[\left(x+\frac{8}{2}\right)^2=(x+4)^2=x^2+8x+16\] Having done this, the \(x\) and \(x^2\) terms are correct!
However, comparing the equations we see that we need to subtract \(4\) from our new expression for it to be equal to the original:
\[\begin{align}
x^2+8x+12 &= x^2+8x+(164)
\\[3pt]&= (x^2+8x+16)4
\\[8pt]&=(x+4)^24
\end{align}\]
Full worked example 1: nice values\[\begin{align} x^2+4x5 &= 0 \\[3pt](x+2)^2 &= x^2+4x+4 \\[3pt]x^2+4x+49 &= x^2+4x5 \\[3pt]x^2+4x+49 &= 0 \\[3pt](x+2)^29 &= 0 \\[3pt](x+2)^2 &= 9 \\[3pt]x+2 &= \pm\sqrt{9} \\[3pt]x+2 &= \pm3 \\[3pt]x &= \pm32 \\[3pt]x &= 1\hspace{8 pt}\text{or }5 \end{align}\] Full worked example 2: working with fractions\[\begin{align} x^2+3x+2 &= 0 \\[8pt] \left(x+\frac{3}{2}\right)^2 &= x^2+3x+\frac{9}{4} \\[8pt] x^2+3x+\frac{9}{4}\frac{1}{4} &= x^2+3x+2 \\[8pt] x^2+3x+\frac{9}{4}\frac{1}{4} &= 0 \\[8pt] \left(x+\frac{3}{2}\right)^2\frac{1}{4} &= 0 \\[8pt] \left(x+\frac{3}{2}\right)^2 &= \frac{1}{4} \\[8pt] x+\frac{3}{2}=\pm\sqrt{\frac{1}{4}} &= \pm\frac{1}{2} \\[8pt] x &= \pm\frac{1}{2}\frac{3}{2} \\[8pt] x &= 1\hspace{8 pt}\text{or }2 \end{align}\] Full worked example 3: surd answer, negative \(x\) coefficient\[\begin{align} x^24x+2&=0 \\[3pt](x2)^2&=x^24x+4 \\[3pt]x^24x+42&=x^24x+2 \\[3pt]x^24x+42&=0 \\[3pt](x2)^22&=0 \\[3pt](x2)^2&=2 \\[3pt]x2&=\pm\sqrt{2} \\[3pt]x&=2\pm\sqrt{2} \end{align}\] Full worked example 4: no real roots\[\begin{align} \\[3pt]x^2+6x+10&=0 \\[3pt](x+3)^2&=x^2+6x+9 \\[3pt]x^2+6x+9+1&=x^2+6x+10 \\[3pt]x^2+6x+9+1&=0 \\[3pt](x+3)^2+1&=0 \\[3pt](x+3)^2&=1 \\[3pt]x+3&=\pm\sqrt{1}\end{align} \\[3pt] \text{Cannot root negative number so no real roots} \] Unless you are doing further maths, if you get an answer in an exam where you're square rooting a negative number then it's likely you've made a mistake! You could check if there should be real roots by using the discriminant \(b^24ac\). You may notice that in the above examples, the coefficient of \(x^2\) is \(1\). In some questions, you may be told to solve an equation where this isn't the case. Here we'd just divide through by the \(x^2\) coefficient before solving. See example below. Full worked example 5: \(x^2\) coefficient \(\neq1\)\[\begin{align} \\[8pt]3x^2+12x+5&=0 \\[8pt]3\left(x^2+4x+\frac{5}{3}\right)&=0 \\[8pt]x^2+4x+\frac{5}{3}&=0 \\[8pt](x+2)^2&=x^2+4x+4 \\[8pt]x^2+4x+4\frac{7}{3}&=x^2+4x+\frac{5}{3} \\[8pt]x^2+4x+4\frac{7}{3}&=0 \\[8pt](x+2)^2\frac{7}{3}&=0 \\[8pt](x+2)^2&=\frac{7}{3} \\[8pt]x+2&=\pm\sqrt{\frac{7}{3}} \\[8pt]x&=\pm\sqrt{\frac{7}{3}}2 \end{align}\] 