Solve using the Quadratic Equation. 1. \( 6 x_{2}-3 x+1=0 \) 2. \( x_{2}-2 x+10=0 \) 3. \( x_{2} \cdot 4 x+5=0 \)

To solve these quadratic equations using the Quadratic Formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), let's break them down one by one. 1. For the equation \( 6x^2 - 3x + 1 = 0 \): - Here, \( a = 6 \), \( b = -3 \), and \( c = 1 \). - Calculate the discriminant: \( b^2 - 4ac = (-3)^2 - 4(6)(1) = 9 - 24 = -15 \). - Since the discriminant is negative, there are no real solutions; the solutions are complex: \( x = \frac{3 \pm i\sqrt{15}}{12} \). 2. For \( x^2 - 2x + 10 = 0 \): - Here, \( a = 1 \), \( b = -2 \), and \( c = 10 \). - Calculate the discriminant: \( b^2 - 4ac = (-2)^2 - 4(1)(10) = 4 - 40 = -36 \). - Again, the discriminant is negative, indicating no real solutions; the solutions are complex too: \( x = \frac{2 \pm 6i}{2} = 1 \pm 3i \). 3. For the equation \( 4x^2 + 5 = 0 \): - Rearranging gives us \( 4x^2 = -5 \). - Thus, \( x^2 = -\frac{5}{4} \). - Taking square roots provides the solutions: \( x = \pm \frac{\sqrt{5}}{2}i \). So the final results are: 1. \( x = \frac{3 \pm i\sqrt{15}}{12} \), 2. \( x = 1 \pm 3i \), 3. \( x = \pm \frac{\sqrt{5}}{2}i \).