(p-1)*x^2+(2p+3)*x+p=0

Предыдущий
вопрос Следующий
вопрос

To solve this quadratic equation, we can use the quadratic formula:

Given a quadratic equation in the form ax^2 + bx + c = 0, the solutions for x are given by the formula:
x = (-b ± sqrt(b^2 - 4ac)) / 2a

In this case, our quadratic equation is in the form of (p-1)x^2 + (2p+3)x + p = 0. So, a = p-1, b = 2p+3, and c = p.

Now we plug these values into the quadratic formula:
x = (-(2p+3) ± sqrt((2p+3)^2 - 4(p-1)(p))) / 2(p-1)

Expanding the terms and simplifying gives:
x = (-2p-3 ± sqrt(4p^2 + 12p + 9 - 4(p^2 - p))) / 2(p-1)
x = (-2p-3 ± sqrt(4p^2 + 12p + 9 - 4p^2 + 4p)) / 2(p-1)
x = (-2p-3 ± sqrt(16p + 9)) / 2(p-1)
x = (-2p-3 ± sqrt(16p + 9)) / 2p - 2

Therefore, the solutions for x are (-2p-3 + sqrt(16p + 9)) / 2p - 2 and (-2p-3 - sqrt(16p + 9)) / 2p - 2.