F(n)=(3n2-4n)2-19(3n2-4n)+60

Solution for F(n)=(3n2-4n)2-19(3n2-4n)+60 equation:

(F)=(3F^2-4F)2-19(3F^2-4F)+60

We move all terms to the left:

(F)-((3F^2-4F)2-19(3F^2-4F)+60)=0


We calculate terms in parentheses: -((3F^2-4F)2-19(3F^2-4F)+60), so:

(3F^2-4F)2-19(3F^2-4F)+60

We multiply parentheses

6F^2-57F^2-8F+76F+60

We add all the numbers together, and all the variables

-51F^2+68F+60

Back to the equation:

-(-51F^2+68F+60)


We get rid of parentheses

51F^2-68F+F-60=0

We add all the numbers together, and all the variables

51F^2-67F-60=0

a = 51; b = -67; c = -60;
Δ = b2-4ac
Δ = -672-4·51·(-60)
Δ = 16729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-67)-\sqrt{16729}}{2*51}=\frac{67-\sqrt{16729}}{102}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-67)+\sqrt{16729}}{2*51}=\frac{67+\sqrt{16729}}{102}
$