(a+2)*x^2+2*(a+2)*x-3=0

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вопрос

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

For an equation in the form of ax^2 + bx + c = 0, the solutions are given by:

x = (-b ± √(b^2 - 4ac)) / 2a

In this case, a = a+2, b = 2(a+2), and c = -3. So, plugging these values into the formula, we get:

x = [-(2(a+2)) ± √((2(a+2))^2 - 4(a+2)(-3))] / 2(a+2)

Simplifying further:

x = [-2a - 4 ± √(4(a^2 + 4a + 4) + 12(a+2))] / 2(a+2)
x = [-2a - 4 ± √(4a^2 + 16a + 16 + 12a + 24)] / 2(a+2)
x = [-2a - 4 ± √(4a^2 + 28a + 40)] / 2(a+2)
x = [-2a - 4 ± √((2a + 4)(2a + 10))] / 2(a+2)
x = [-2a - 4 ± (2a + 4)√(a + 5)] / 2(a+2)
x = [-2a - 4 ± 2(a + 4)√(a + 5)] / 2(a+2)
x = [(-2a - 4 ± 2(a + 4)√(a + 5)] / 2(a+2)

Therefore, the solutions to the quadratic equation are:

x = (-2a - 4 + 2(a + 4)√(a + 5)) / 2(a+2)
x = (-2a - 4 - 2(a + 4)√(a + 5)) / 2(a+2)