Evaluate: 10x-3(x^2-16)=0

Expression: $10x-3\left( {x}^{2}-16 \right)=0$

Distribute $-3$ through the parentheses

$10x-3{x}^{2}+48=0$

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

$\begin{align*}&\begin{array} { l }x_1=-\frac{ 8 }{ 3 },& x_2=6\end{array} \\&\begin{array} { l }x_1\approx-2.66667,& x_2=6\end{array}\end{align*}$