Sure, let's solve the equation step-by-step.
We start with the equation:
[tex]$(3x + 2)(x - 3) = 7x - 1$[/tex]
First, we need to expand the left-hand side:
[tex]$(3x + 2)(x - 3) = 3x(x - 3) + 2(x - 3)$[/tex]
[tex]$= 3x^2 - 9x + 2x - 6$[/tex]
[tex]$= 3x^2 - 7x - 6$[/tex]
So, the equation can be rewritten as:
[tex]$3x^2 - 7x - 6 = 7x - 1$[/tex]
Next, we need to bring all the terms to one side of the equation:
[tex]$3x^2 - 7x - 6 - 7x + 1 = 0$[/tex]
[tex]$3x^2 - 14x - 5 = 0$[/tex]
Now, we solve the quadratic equation [tex]\(3x^2 - 14x - 5 = 0\)[/tex]. The quadratic formula is given by:
[tex]$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$[/tex]
For our quadratic equation, [tex]\(a = 3\)[/tex], [tex]\(b = -14\)[/tex], and [tex]\(c = -5\)[/tex]. Plugging in these values gives:
[tex]$x = \frac{-(-14) \pm \sqrt{(-14)^2 - 4 \cdot 3 \cdot (-5)}}{2 \cdot 3}$[/tex]
[tex]$x = \frac{14 \pm \sqrt{196 + 60}}{6}$[/tex]
[tex]$x = \frac{14 \pm \sqrt{256}}{6}$[/tex]
[tex]$x = \frac{14 \pm 16}{6}$[/tex]
This gives us two solutions:
[tex]$x = \frac{14 + 16}{6} = \frac{30}{6} = 5$[/tex]
[tex]$x = \frac{14 - 16}{6} = \frac{-2}{6} = -\frac{1}{3}$[/tex]
So, the solutions to the equation [tex]\((3x + 2)(x - 3) = 7x - 1\)[/tex] are:
[tex]$x = 5 \quad \text{and} \quad x = -\frac{1}{3}$[/tex]
