Question

quadratic by factoring.
$3x^2 - 20x - 13 = -6$

Explanation:

Step1: Rearrange the equation

First, we need to set the equation to standard quadratic form \(ax^2 + bx + c = 0\). So, add 6 to both sides of the equation \(3x^2 - 20x - 13 = -6\):
\(3x^2 - 20x - 13 + 6 = 0\)
Simplify to get:
\(3x^2 - 20x - 7 = 0\)

Step2: Factor the quadratic

We need to find two numbers that multiply to \(3\times(-7)= -21\) and add up to \(-20\). The numbers are \(-21\) and \(1\) because \(-21\times1 = -21\) and \(-21 + 1 = -20\).
Now, rewrite the middle term using these numbers:
\(3x^2 - 21x + x - 7 = 0\)
Group the terms:
\((3x^2 - 21x) + (x - 7) = 0\)
Factor out the greatest common factor from each group:
\(3x(x - 7) + 1(x - 7) = 0\)
Now, factor out the common binomial factor \((x - 7)\):
\((3x + 1)(x - 7) = 0\)

Step3: Solve for x

Set each factor equal to zero and solve for \(x\):

• For \(3x + 1 = 0\):

Subtract 1 from both sides: \(3x = -1\)
Divide both sides by 3: \(x = -\frac{1}{3}\)

• For \(x - 7 = 0\):

Add 7 to both sides: \(x = 7\)

Answer:

\(x = -\frac{1}{3}\) or \(x = 7\)