Question

Z^2+(7-4)z+3(7+11)=0

Answer 1

z²+(7-4)z+3(7+11)=0

z²+3z+3*18=0

z²+3z+54=0

D = (3)²-4*1*54 = 9-216 = -207, D < 0

Нет корней.

Ответ: нет корней

Answer 2

To solve the quadratic equation Z^2 + (7 - 4)z + 3(7 + 11) = 0, we need to simplify the equation and then apply the quadratic formula.

First, simplify the equation: Z^2 + 3z + 3(18) = 0 Z^2 + 3z + 54 = 0

Now we can use the quadratic formula, which states that for an equation in the form ax^2 + bx + c = 0, the solutions are given by:

x = (-b ± √(b^2 - 4ac)) / (2a)

In our case, a = 1, b = 3, and c = 54. Substituting these values into the quadratic formula, we get:

Z = (-3 ± √(3^2 - 4(1)(54))) / (2(1)) Z = (-3 ± √(9 - 216)) / 2 Z = (-3 ± √(-207)) / 2

Since the discriminant (√(b^2 - 4ac)) is negative (-207 is negative), the solutions will involve complex numbers.

Using the imaginary unit, denoted by "i," where i^2 = -1, we can simplify the square root of -207:

√(-207) = √(207) √(-1) = √(207)i

Now we can rewrite the solutions:

Z = (-3 ± √(-207)) / 2 Z = (-3 ± √(207)i) / 2

Thus, the solutions to the quadratic equation Z^2 + (7 - 4)z + 3(7 + 11) = 0 are:

Z = (-3 + √(207)i) / 2 Z = (-3 - √(207)i) / 2