Answer
True.
The values of x are $10\text{ and }-7$.
Work Step by Step
Consider the equation.
$x\left( x-3 \right)=70$
Apply the distributive property.
${{x}^{2}}-3x=70$
Apply the addition principle:
$\begin{align}
& {{x}^{2}}-3x-70=70-70 \\
& {{x}^{2}}-3x-70=0 \\
\end{align}$
Factor:
$\begin{align}
& {{x}^{2}}-10x+7x-70=0 \\
& x\left( x-10 \right)+7\left( x-10 \right)=0 \\
& \left( x-10 \right)\left( x+7 \right)=0
\end{align}$
Apply the principle of zero products.
$\begin{align}
& x-10=0\text{ or }x+7=0 \\
& x=10\text{ or }x=-7
\end{align}$
Therefore, the values of x are 10 and $-7$ respectively.
Check:
Substitute 10 for $x$ in the given equation $x\left( x-3 \right)=70$.
$\begin{align}
10\left( 10-3 \right)\overset{?}{\mathop{=}}\,70 & \\
10\left( 7 \right)\overset{?}{\mathop{=}}\,70 & \\
70=70 & \\
\end{align}$
Substitute $-7$ for $x$ in the given equation $x\left( x-3 \right)=70$.
$\begin{align}
-7\left( -7-3 \right)\overset{?}{\mathop{=}}\,70 & \\
-7\left( -10 \right)\overset{?}{\mathop{=}}\,70 & \\
70=70 & \\
\end{align}$
It is also true.
