Simplifying quadratic formula

I must have forgotten some rule... trying to figure out how to simplify this... 

$x=\frac{4\left(a+b\right)±\sqrt{16\left(a+b\right)^2-48ab}}{24}$

Simplifies to: $x=\frac{\left(a+b\right)±\sqrt{a^2-ab+b^2}}{6}$

$\begin{array}{|rcll|} \hline x &=& \frac{4\left(a+b\right) \pm \sqrt{16\cdot \left(a+b\right)^2-48ab}}{24} \\ &=& \frac{4\left(a+b\right) \pm \sqrt{16\cdot \left(a+b\right)^2-16\cdot 3 \cdot ab}}{24} \\ &=& \frac{4\left(a+b\right) \pm \sqrt{ 16 \cdot[~ \left(a+b\right)^2-3\cdot ab ~]}}{24} \quad & | \quad \sqrt{a\cdot b} = \sqrt{a}\cdot \sqrt{ b} \\ &=& \frac{4\left(a+b\right) \pm \sqrt{ 16}\cdot \sqrt{ \left(a+b\right)^2-3ab } }{24} \quad & | \quad \sqrt{ 16}=4\\ &=& \frac{4\left(a+b\right) \pm 4\cdot \sqrt{ \left(a+b\right)^2-3ab } }{24} \\ &=& \frac{4 \cdot[~ \left(a+b\right) \pm \sqrt{ \left(a+b\right)^2-3ab} ~] }{24} \\ &=& \frac{4 \cdot[~ \left(a+b\right) \pm \sqrt{ \left(a+b\right)^2-3ab} ~] }{4\cdot 6} \\ &=& \frac{ \left(a+b\right) \pm \sqrt{ \left(a+b\right)^2-3ab} } { 6} \quad & | \quad (a+b)^2 = a^2 + 2ab + b^2\\ &=& \frac{ \left(a+b\right) \pm \sqrt{ a^2 + 2ab + b^2-3ab} } { 6} \\ &=& \frac{ \left(a+b\right) \pm \sqrt{ a^2 + 2ab-3ab + b^2} } { 6} \quad & | \quad 2ab-3ab = ab(2-3)= ab(-1) = -ab\\ &=& \frac{ \left(a+b\right) \pm \sqrt{ a^2 -ab + b^2} } { 6} \\ \hline \end{array}$