```Question 366375
11(x-2) + (x-5) = (x+2)(x-6)
This is a quadratic equation because there will be an {{{x^2}}} term on the right side if you multiply it out. Most of the time, with quadratic equations, you want the equation in the form: {{{ax^2 + bx + c = 0}}}. This involves simplifying each side of the equation and then making one side of the equation zero.<br>
And this is exactly what we will be doing with your equation. We start by simplifying (i.e. go through the order of operations (aka PEMDAS)) on each side of the equation:
{{{11x - 22 + (x - 5) = x^2 -4x - 12}}}
Combining like terms on the left side:
{{{12x - 27 = x^2 -4x -12}}}
Each side is now simplified. Next we want one side to be zero. Since the solution will be easier with {{{x^2}}} than with {{{-x^2}}}, I will make the left side zero by subtracting 12x from each side and adding 27 to each side:
{{{0 = x^2 -16x + 15}}}
Now we have the desired form. From here we wither factor or use the Quadratic Formula. This expression factors fairly easily:
0 = (x - 15)(x - 1)
From the Zero Product Property we know that this product can be zero <i>only</i> if one of the factors is zero. So:
x - 15 = 0 or x - 1 = 0
Solving each of these we get:
x = 15 or x = 1```