(6x-1)^2-3x(9x-2)=(3x+4)^2

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вопрос

Expanding both sides of the equation:

Left side:
(6x-1)^2 - 3x(9x-2)
= (6x-1)(6x-1) - 27x^2 + 6x
= 36x^2 - 6x - 6x + 1 - 27x^2 + 6x
= 9x^2 + 1

Right side:
(3x+4)^2
= (3x+4)(3x+4)
= 9x^2 + 12x + 12x + 16
= 9x^2 + 24x + 16

Therefore, the equation simplifies to:
9x^2 + 1 = 9x^2 + 24x + 16

Since the x^2 terms are equal, we can solve for x:

1 = 24x + 16
1 - 16 = 24x
-15 = 24x
x = -15/24
x = -5/8

So, the solution to the equation is x = -5/8.