Given
[tex]4a^2-48a+52=0[/tex]
Solution
[tex]\begin{gathered} 4a^2-48a+52=0 \\ So\text{ divide both sides by 4} \\ \frac{4a^2}{4}-\frac{48a}{4}+\frac{52}{4}=0 \\ which\text{ gives } \\ a^2-12a+13=0 \\ Keep\text{ a on LHS} \\ a^2-12a=-13 \\ Take\text{ the half of coefficient of a and square it} \\ (-\frac{12}{2})^2=36 \\ \\ rewrite\text{ as perfect square} \\ \\ (a-6)^2=-13+36 \\ (a-6)^2=23 \\ \end{gathered}[/tex]
[tex]\begin{gathered} a-6=\pm\sqrt{23} \\ a=6\pm\sqrt{23} \end{gathered}[/tex]
[tex]\begin{gathered} a=10.79583\text{ ,}a=1.20416 \\ or \\ a=6+\sqrt{23},\:a=6-\sqrt{23}\quad \\ \\ \end{gathered}[/tex]
