F(2a)=-2(2a)^2+2(2a)

Solution for F(2a)=-2(2a)^2+2(2a) equation:

(2F)=-2(2F)^2+2(2F)

We move all terms to the left:

(2F)-(-2(2F)^2+2(2F))=0

determiningTheFunctionDomain
-(-22F^2+22F)+2F=0

We get rid of parentheses

22F^2-22F+2F=0

We add all the numbers together, and all the variables

22F^2-20F=0

a = 22; b = -20; c = 0;
Δ = b2-4ac
Δ = -202-4·22·0
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-20}{2*22}=\frac{0}{44}
=0
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+20}{2*22}=\frac{40}{44}
=10/11
$