(1,8 -0,3у) *(2у+9)=0

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вопрос

To solve this equation, we can distribute the first term over the second term:

(1,8 -0,3y)(2y+9) = 0
1,8(2y) + 1,89 - 0,3y(2y) - 0,3y*9 = 0

Now simplify the equation:

3,6y + 16,2 - 0,6y^2 - 2,7y = 0
-0,6y^2 + 0.9y + 16.2 = 0

Now we have a quadratic equation. To solve for y, we can use the quadratic formula:

y = (-b ± √(b^2-4ac)) / 2a

In this case, a = -0,6, b = 0,9, and c = 16,2. Plugging those values into the equation, we get:

y = (-0,9 ± √(0,9^2-4(-0,6)16,2)) / 2*(-0,6)
y = (-0.9 ± √(0.81 + 38.88)) / -1.2
y = (-0.9 ± √39.69) / -1.2
y = (-0.9 ± 6.3) / -1.2

So, the solutions for y are:
y = (5.4 / -1.2) = -4.5
y = (-7.2 / -1.2) = 6

Therefore, the solutions for y are y = -4.5 and y = 6.