(р+2)х в квадрате +(р+2)х+2=0

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To solve this quadratic equation, let's first substitute (р+2)х as a variable, let's call it y:

y^2 + y + 2 = 0

Now we can solve the quadratic equation for y using the quadratic formula:

y = (-b ± √(b^2 - 4ac)) / 2a

In this case, a = 1, b = 1, and c = 2. Plugging these values into the formula, we get:

y = (-1 ± √(1^2 - 412)) / 2*1
y = (-1 ± √(1 - 8)) / 2
y = (-1 ± √(-7)) / 2

Since the square root of a negative number is not a real number, the equation has no real solutions.