Реши уравнение

1. x^2+3x-18=0

2. x^2+6=5x

3. 2x^2-3x больше 0

1. x^2+3x-18=0

D = b^2 - 4ac = 3^2 - 4*1 * (-18) = 72 + 9 = 81

x1 = - b + √D / 2a = - 3 + 9 / 2 = 3

x2 = - b - √D / 2a = - 3-9/2 = - 6

2. x^2+6=5x

x^2 + 6 - 5x = 0

x^2 - 5x + 6 = 0

D = b^2 - 4ac = (-5) ^2 - 4*1*6 = 25 - 24 = 1

x1 = - b + √D / 2a = 5 + 1/2 = 3

x2 = - b - √D / 2a = 5 - 1/2 = 2

3. 2x^2-3x больше 0

x (2x-3) больше 0

x = 0 или 2x-3 = 0

2x = 3

x = 3/2

x = 1,5

X^2+3x-18=0

D=9-4*1 * (-18) = 9+72=81

x1,2=-3+9/2 = - 6, 3

в ответе просто записать корни как:

х1=3

х2=-6