The problem asks us to find the solution set for each of the given quadratic equations using the quadratic formula. The equations are: a) $2x^2 - 7x + 1 = 0$ b) $3x^2 = -9x + 2$ c) $x^2 - 4x = 3$ d) $x^2 = x + 5$

1. Problem Description

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

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

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

c) $x^2 - 4x = 3$

d) $x^2 = x + 5$

2. Solution Steps

The quadratic formula is given by:

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

a) $2x^2 - 7x + 1 = 0$. Here, $a = 2$, $b = -7$, and $c = 1$.

$x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(2)(1)}}{2(2)} = \frac{7 \pm \sqrt{49 - 8}}{4} = \frac{7 \pm \sqrt{41}}{4}$.

The solution set is $\{\frac{7 + \sqrt{41}}{4}, \frac{7 - \sqrt{41}}{4}\}$.

b) $3x^2 = -9x + 2$. First, rewrite the equation in standard form: $3x^2 + 9x - 2 = 0$. Here, $a = 3$, $b = 9$, and $c = -2$.

$x = \frac{-9 \pm \sqrt{9^2 - 4(3)(-2)}}{2(3)} = \frac{-9 \pm \sqrt{81 + 24}}{6} = \frac{-9 \pm \sqrt{105}}{6}$.

The solution set is $\{\frac{-9 + \sqrt{105}}{6}, \frac{-9 - \sqrt{105}}{6}\}$.

c) $x^2 - 4x = 3$. First, rewrite the equation in standard form: $x^2 - 4x - 3 = 0$. Here, $a = 1$, $b = -4$, and $c = -3$.

$x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-3)}}{2(1)} = \frac{4 \pm \sqrt{16 + 12}}{2} = \frac{4 \pm \sqrt{28}}{2} = \frac{4 \pm 2\sqrt{7}}{2} = 2 \pm \sqrt{7}$.

The solution set is $\{2 + \sqrt{7}, 2 - \sqrt{7}\}$.

d) $x^2 = x + 5$. First, rewrite the equation in standard form: $x^2 - x - 5 = 0$. Here, $a = 1$, $b = -1$, and $c = -5$.

$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-5)}}{2(1)} = \frac{1 \pm \sqrt{1 + 20}}{2} = \frac{1 \pm \sqrt{21}}{2}$.

The solution set is $\{\frac{1 + \sqrt{21}}{2}, \frac{1 - \sqrt{21}}{2}\}$.

3. Final Answer

a) $\{\frac{7 + \sqrt{41}}{4}, \frac{7 - \sqrt{41}}{4}\}$

b) $\{\frac{-9 + \sqrt{105}}{6}, \frac{-9 - \sqrt{105}}{6}\}$

c) $\{2 + \sqrt{7}, 2 - \sqrt{7}\}$

d) $\{\frac{1 + \sqrt{21}}{2}, \frac{1 - \sqrt{21}}{2}\}$