Плез решение
x^2-18x^2+32=0
(6x-1)*(x+1)=20

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вопрос

To solve the equation given:

Factor the equation x^2 - 18x + 32 = 0:
x^2 - 18x + 32 = (x - 16)(x - 2) = 0

Set each factor to zero:
x - 16 = 0 or x - 2 = 0

Solve for x in each case:
x = 16 or x = 2

Therefore, the solutions to the equation x^2 - 18x + 32 = 0 are x = 16 and x = 2.

To solve the equation (6x-1)*(x+1) = 20:

Expand the equation:
6x^2 + 6x - x - 1 = 20
6x^2 + 5x - 1 = 20

Rearrange terms to get a quadratic equation in standard form:
6x^2 + 5x - 21 = 0

Factor the quadratic equation:
6x^2 + 14x - 9x - 21 = 0
2x(3x + 7) - 3(3x + 7) = 0
(2x - 3)(3x + 7) = 0

Set each factor to zero and solve for x:
2x - 3 = 0 or 3x + 7 = 0
2x = 3 or 3x = -7
x = 3/2 or x = -7/3

Therefore, the solutions to the equation (6x-1)*(x+1) = 20 are x = 3/2 and x = -7/3.