Respuesta :

Lets take this question as:

[tex]\left(3x-3\right)\lbrack6(x-10)\rbrack[/tex]

First, we distribute the parenthesis (2nd one):

[tex]\begin{gathered} \left(3x-3\right)\lbrack6(x-10)\rbrack \\ =\left(3x-3\right)\lbrack6x-60\rbrack \end{gathered}[/tex]

Multiply it out:

[tex]\begin{gathered} \left(3x-3\right)(6x-60) \\ =(3x)(6x)-(3x)(60)-(3)(6x)+(3)(60) \\ =18x^2-180x-18x+180 \end{gathered}[/tex]

Now, we combine like terms and factor common term out:

[tex]\begin{gathered} 18x^2-180x-18x+180 \\ =18x^2-198x+180 \\ =18(x^2-11x+10) \end{gathered}[/tex]

Let's equate this to 0 and solve for x by factoring:

[tex]\begin{gathered} 18(x^2-11x+10)=0 \\ x^2-11x+10=0 \\ (x-1)(x-10)=0 \\ x=1,10 \end{gathered}[/tex]

The solution is:

[tex]x=1\text{ and x=10}[/tex]