What are the solutions of the equation x^4 - 7x^3 + 23x^3 - 29x - 60 = 0 are shown. What are the nonreal solutions to the equation?

Without having the correct equation, it's not possible to accurately provide the nonreal solutions. However, nonreal solutions for an equation generally occur when the discriminant in the quadratic formula is less than zero, leading to a complex solution.
The given equation x^4 - 7x^3 + 23x^3 - 29x - 60 = 0 is a polynomial equation of fourth degree. The nonreal solutions of this equation are also known as complex solutions. The nonreal solutions of the equation cannot be found without further information as, it seems there might be a typo in the equation. However, in general to find nonreal solutions, we apply the quadratic formula -b ± √b² - 4ac / 2a where the discriminant (b^2 - 4ac) is negative.
Let's say we have an equation of the form ax² + bx + c = 0. Here, a, b, and c are coefficients. If 'b² - 4ac' is less than 0, then the roots are nonreal or complex. They are calculated using the quadratic formula -b ± √b² - 4ac / 2a. So, in the context of your question, we would need to apply this process to the equation to find the nonreal solutions.
Please note, this would likely involve simplifying the given equation to a quadratic form and then applying the quadratic formula, looking especially for places where the discriminant results in a negative square root, leading to a complex solution.
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Answer:
A.
Step-by-step explanation:
1) according to the given graph, the real roots are: -1 and 4, it means the given expression can be evaluated using (x+1)(x-4)=x²-3x-4.
2) if to divide the given expression by (x²-3x-4), then
x⁴-7x³+23x²-29x-60=(x²-3x-4)(x²-4x-15), where
3) x=-1;4 - real roots, and
[tex]x=2^+_-i\sqrt{11} \ -nonreal \ roots.[/tex]