To calculate: The solution set of the polynomial equation n3n+2+1=4n−2.

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Greg Snyder · Accepted Answer

Given Information:    The provided polynomial equation is n3n+2+1=4n−2.    Formula Used:    The solution of quadratic equation ax2+bx+c=0 is,    x=−b±b2−4ac2a  Calculation:    Simplify the equation n3n+2+1=4n−2; by multiplying (3n+2)(n—2) on both the sides,    (3n+2)(n—2)[n3n+2+1]=(3n+2)(n—2)[4n−2]    n(n—2)+(3n+2)(n—2)=(3n+2)(4)    n2−2n+3n2−6n+2n−4=12n+8    4n2−18n−12=0    This is a polynomial equation with one side equal to zero.    4n2−18n−12=0    Solve the equation 4n2−18n−12=0,    n=−(−18)±(−18)2−4(4)(−12)2(4)    n=18±324+1928    n=18±21298    n=9±1294    Check at n=9+1294    PSK9+12943(9+1294)+2+149+1294−2    5.0815.26+145.02−2    0.33+11.32    1.33=1.32    The left side value is equal to the right-side value. Thus, the value n=9+1294 is verified.    Check at n=

To calculate: The solution set of the polynomial equation \frac{n}{

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