F(x)=(3x-2x^2)5

Solution for F(x)=(3x-2x^2)5 equation:

(F)=(3F-2F^2)5

We move all terms to the left:

(F)-((3F-2F^2)5)=0


We calculate terms in parentheses: -((3F-2F^2)5), so:

(3F-2F^2)5

We multiply parentheses

-10F^2+15F

Back to the equation:

-(-10F^2+15F)


We get rid of parentheses

10F^2-15F+F=0

We add all the numbers together, and all the variables

10F^2-14F=0

a = 10; b = -14; c = 0;
Δ = b2-4ac
Δ = -142-4·10·0
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-14}{2*10}=\frac{0}{20}
=0
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+14}{2*10}=\frac{28}{20}
=1+2/5
$