Having obtained:
[tex]x^2+ \frac{b}{a}x+( \frac{b}{2a} )^2+ \frac{c}{a}-( \frac{b}{2a} )^2=0 [/tex]
the next thing to do is collect the first three terms and write them as the square of a binomial, and also collect the last two terms, writing each of them with a denominator equal to 4a^2
We have:
[tex](x+\frac{b}{2a} )^2+( \frac{4ac}{4a^2}-\frac{b^2}{4a^2} )=0[/tex].
Then, we take [tex]( \frac{4ac}{4a^2}-\frac{b^2}{4a^2} )[/tex] to the right side and write these two terms as one:
[tex](x+\frac{b}{2a} )^2=\frac{b^2-4ac}{4a^2} [/tex].
Next, we take the square root of both sides, which has been shown in the solution.
Next, we have to take b/2a to the right hand side as -(b/2a), and removing the square in the denominator of the right hand side expression:
[tex]x=-\frac{b}{2a}\pm \frac{\sqrt{b^2-4ac}}{2a}}[/tex].
Answer: the steps to complete in the boxes are :
[tex](x+\frac{b}{2a} )^2+( \frac{4ac}{4a^2}-\frac{b^2}{4a^2} )=0[/tex].
[tex](x+\frac{b}{2a} )^2=\frac{b^2-4ac}{4a^2} [/tex].
[tex]x=-\frac{b}{2a}\pm \frac{\sqrt{b^2-4ac}}{2a}}[/tex].
