To solve the equation 3x + 5 = 0, we need to isolate the variable x on one side of the equation. We can do this by subtracting 5 from both sides of the equation, which gives us 3x = -5. Then, we can divide both sides of the equation by 3, which gives us x = -5/3.

To solve the quadratic equation x^2 + 4x + 3 = 0, we can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a. In this case, a = 1, b = 4, and c = 3. Plugging these values into the formula, we get x = (-4 ± √(4^2 - 4(1)(3))) / 2(1). Simplifying this expression, we get x = (-4 ± √(16 - 12)) / 2. Further simplifying, we get x = (-4 ± √4) / 2. Finally, we get x = (-4 ± 2) / 2. This gives us two solutions: x = -1 and x = -3.

To solve the equation 2x - 7 = 1, we need to isolate the variable x on one side of the equation. We can do this by adding 7 to both sides of the equation, which gives us 2x = 8. Then, we can divide both sides of the equation by 2, which gives us x = 4.

To solve the quadratic equation x^2 - 5x + 6 = 0, we can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a. In this case, a = 1, b = -5, and c = 6. Plugging these values into the formula, we get x = (-(-5) ± √((-5)^2 - 4(1)(6))) / 2(1). Simplifying this expression, we get x = (5 ± √(25 - 24)) / 2. Further simplifying, we get x = (5 ± √1) / 2. Finally, we get x = (5 ± 1) / 2. This gives us two solutions: x = 2 and x = 3.

To solve the equation 4x + 2 = 10, we need to isolate the variable x on one side of the equation. We can do this by subtracting 2 from both sides of the equation, which gives us 4x = 8. Then, we can divide both sides of the equation by 4, which gives us x = 2.

To solve the quadratic equation x^2 + 2x - 3 = 0, we can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a. In this case, a = 1, b = 2, and c = -3. Plugging these values into the formula, we get x = (-2 ± √(2^2 - 4(1)(-3))) / 2(1). Simplifying this expression, we get x = (-2 ± √(4 + 12)) / 2. Further simplifying, we get x = (-2 ± √16) / 2. Finally, we get x = (-2 ± 4) / 2. This gives us two solutions: x = 1 and x = -3.

To solve the equation 5x - 3 = 7, we need to isolate the variable x on one side of the equation. We can do this by adding 3 to both sides of the equation, which gives us 5x = 10. Then, we can divide both sides of the equation by 5, which gives us x = 2.

To solve the quadratic equation x^2 - 4x + 4 = 0, we can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a. In this case, a = 1, b = -4, and c = 4. Plugging these values into the formula, we get x = (-(-4) ± √((-4)^2 - 4(1)(4))) / 2(1). Simplifying this expression, we get x = (4 ± √(16 - 16)) / 2. Further simplifying, we get x = (4 ± √0) / 2. Finally, we get x = (4 ± 0) / 2. This gives us two solutions: x = 2 and x = 2.

To solve the equation 6x + 1 = 13, we need to isolate the variable x on one side of the equation. We can do this by subtracting 1 from both sides of the equation, which gives us 6x = 12. Then, we can divide both sides of the equation by 6, which gives us x = 2.

To solve the quadratic equation x^2 + 6x + 9 = 0, we can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a. In this case, a = 1, b = 6, and c = 9. Plugging these values into the formula, we get x = (-6 ± √(6^2 - 4(1)(9))) / 2(1). Simplifying this expression, we get x = (-6 ± √(36 - 36)) / 2. Further simplifying, we get x = (-6 ± √0) / 2. Finally, we get x = (-6 ± 0) / 2. This gives us two solutions: x = -3 and x = -3.

To solve the equation 7x - 4 = 10, we need to isolate the variable x on one side of the equation. We can do this by adding 4 to both sides of the equation, which gives us 7x = 14. Then, we can divide both sides of the equation by 7, which gives us x = 2.

To solve the quadratic equation x^2 - 2x - 3 = 0, we can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a. In this case, a = 1, b = -2, and c = -3. Plugging these values into the formula, we get x = (-(-2) ± √((-2)^2 - 4(1)(-3))) / 2(1). Simplifying this expression, we get x = (2 ± √(4 + 12)) / 2. Further simplifying, we get x = (2 ± √16) / 2. Finally, we get x = (2 ± 4) / 2. This gives us two solutions: x = 1 and x = -3.

To solve the equation 8x + 5 = 21, we need to isolate the variable x on one side of the equation. We can do this by subtracting 5 from both sides of the equation, which gives us 8x = 16. Then, we can divide both sides of the equation by 8, which gives us x = 2.

To solve the quadratic equation x^2 + 4x + 2 = 0, we can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a. In this case, a = 1, b = 4, and c = 2. Plugging these values into the formula, we get x = (-4 ± √(4^2 - 4(1)(2))) / 2(1). Simplifying this expression, we get x = (-4 ± √(16 - 8)) / 2. Further simplifying, we get x = (-4 ± √8) / 2. Finally, we get x = (-4 ± 2√2) / 2. This gives us two solutions: x = -2 - √2 and x = -2 + √2.

To solve the equation 9x - 6 = 15, we need to isolate the variable x on one side of the equation. We can do this by adding 6 to both sides of the equation, which gives us 9x = 21. Then, we can divide both sides of the equation by 9, which gives us x = 3.