Question

2 X 1) x² + 4x – 5 = 0; 3) x² + 2x - 15 = 0; - 5) x² - 6x +3=0; ​

Answer 1

1) x1=-5,x2=1
3)x1=-5,x3=3
5)x1=3-(корень)6 x2=3+(корень)6
Но это если надо решить квадратное уравнение !!!

Answer 2

Let's solve the quadratic equations you provided one by one:

1. x² + 4x - 5 = 0

To solve this equation, we can factorize it or use the quadratic formula. Let's use the quadratic formula:

The quadratic formula states that for an equation in the form ax² + bx + c = 0, the solutions for x can be found using the formula:

x = (-b ± √(b² - 4ac)) / (2a)

For our equation x² + 4x - 5 = 0, we have a = 1, b = 4, and c = -5. Substituting these values into the quadratic formula:

x = (-4 ± √(4² - 4 * 1 * -5)) / (2 * 1) x = (-4 ± √(16 + 20)) / 2 x = (-4 ± √36) / 2 x = (-4 ± 6) / 2

This gives us two possible solutions: x₁ = (-4 + 6) / 2 = 1 x₂ = (-4 - 6) / 2 = -5

Therefore, the solutions to the equation x² + 4x - 5 = 0 are x = 1 and x = -5.

2. x² + 2x - 15 = 0

Similarly, let's use the quadratic formula to solve this equation:

a = 1, b = 2, and c = -15

x = (-2 ± √(2² - 4 * 1 * -15)) / (2 * 1) x = (-2 ± √(4 + 60)) / 2 x = (-2 ± √64) / 2 x = (-2 ± 8) / 2

This gives us two possible solutions: x₁ = (-2 + 8) / 2 = 3 x₂ = (-2 - 8) / 2 = -5

Therefore, the solutions to the equation x² + 2x - 15 = 0 are x = 3 and x = -5.

3. x² - 6x + 3 = 0

Again, let's use the quadratic formula to solve this equation:

a = 1, b = -6, and c = 3

x = (-(-6) ± √((-6)² - 4 * 1 * 3)) / (2 * 1) x = (6 ± √(36 - 12)) / 2 x = (6 ± √24) / 2 x = (6 ± 2√6) / 2 x = 3 ± √6

This gives us two possible solutions: x₁ = 3 + √6 x₂ = 3 - √6

Therefore, the solutions to the equation x² - 6x + 3 = 0 are x = 3 + √6 and x = 3 - √6.