$10x-3{x}^{2}+48=0$
Move the constant to the right-hand side and change its sign$10x-3{x}^{2}=-48$
Use the commutative property to reorder the terms$-3{x}^{2}+10x=-48$
Divide both sides of the equation by $-3$${x}^{2}-\frac{ 10 }{ 3 }x=16$
To complete the square, the same value needs to be added to both sides${x}^{2}-\frac{ 10 }{ 3 }x+?=16+?$
To complete the square ${x}^{2}-\frac{ 10 }{ 3 }x+\frac{ 25 }{ 9 }={\left( x-\frac{ 5 }{ 3 } \right)}^{2}$ add $\frac{ 25 }{ 9 }$ to the expression${x}^{2}-\frac{ 10 }{ 3 }x+\frac{ 25 }{ 9 }=16+?$
Since $\frac{ 25 }{ 9 }$ was added to the left-hand side, also add $\frac{ 25 }{ 9 }$ to the right-hand side${x}^{2}-\frac{ 10 }{ 3 }x+\frac{ 25 }{ 9 }=16+\frac{ 25 }{ 9 }$
Use ${a}^{2}-2ab+{b}^{2}={\left( a-b \right)}^{2}$ to factor the expression${\left( x-\frac{ 5 }{ 3 } \right)}^{2}=16+\frac{ 25 }{ 9 }$
Calculate the sum${\left( x-\frac{ 5 }{ 3 } \right)}^{2}=\frac{ 169 }{ 9 }$
Solve the equation for $x$$\begin{array} { l }x=-\frac{ 8 }{ 3 },\\x=6\end{array}$
The equation has $2$ solutions$\begin{align*}&\begin{array} { l }x_1=-\frac{ 8 }{ 3 },& x_2=6\end{array} \\&\begin{array} { l }x_1\approx-2.66667,& x_2=6\end{array}\end{align*}$