(3x-1)(2x-2)=(x-4)^2+7

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вопрос

Expanding the left side of the equation:

(3x-1)(2x-2)
= 3x(2x) +3x(-2) -1(2x) -1(-2)
= 6x^2 - 6x - 2x + 2
= 6x^2 - 8x + 2

Expanding the right side of the equation:

(x-4)^2 + 7
= (x-4)(x-4) + 7
= x(x) -4(x) -4(x) + 4(4) + 7
= x^2 - 4x - 4x + 16 + 7
= x^2 - 8x + 23

Setting the two sides equal to each other:

6x^2 - 8x + 2 = x^2 - 8x + 23
Subtract x^2 and -8x from both sides:
5x^2 + 10 = 23
Subtracting 10 from both sides:
5x^2 = 13
Dividing by 5:
x^2 = 13/5
Taking square root of both sides:
x = ±√(13/5)

Therefore, the solutions are:
x = √(13/5) and x = -√(13/5)