The problem asks to find the solution set of the given quadratic equations using the quadratic formula. The quadratic equations are: a) $2x^2 - 7x + 1 = 0$ b) $3x^2 = -9x + 2$

1. Problem Description

The problem asks to find the solution set of the given quadratic equations using the quadratic formula. The quadratic equations are:

a) $2x^2 - 7x + 1 = 0$

b) $3x^2 = -9x + 2$

2. Solution Steps

a) $2x^2 - 7x + 1 = 0$

We use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 2$, $b = -7$, and $c = 1$.

Substituting these values into the formula:

$x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(2)(1)}}{2(2)}$

$x = \frac{7 \pm \sqrt{49 - 8}}{4}$

$x = \frac{7 \pm \sqrt{41}}{4}$

So, the two solutions are $x = \frac{7 + \sqrt{41}}{4}$ and $x = \frac{7 - \sqrt{41}}{4}$.

b) $3x^2 = -9x + 2$

First, rewrite the equation in the standard quadratic form:

$3x^2 + 9x - 2 = 0$

Here, $a = 3$, $b = 9$, and $c = -2$.

Substituting these values into the quadratic formula:

$x = \frac{-9 \pm \sqrt{9^2 - 4(3)(-2)}}{2(3)}$

$x = \frac{-9 \pm \sqrt{81 + 24}}{6}$

$x = \frac{-9 \pm \sqrt{105}}{6}$

So, the two solutions are $x = \frac{-9 + \sqrt{105}}{6}$ and $x = \frac{-9 - \sqrt{105}}{6}$.

3. Final Answer

a) The solution set for $2x^2 - 7x + 1 = 0$ is $\{\frac{7 + \sqrt{41}}{4}, \frac{7 - \sqrt{41}}{4}\}$.

b) The solution set for $3x^2 = -9x + 2$ is $\{\frac{-9 + \sqrt{105}}{6}, \frac{-9 - \sqrt{105}}{6}\}$.