F(5a)=2(5a)^2

Solution for F(5a)=2(5a)^2 equation:

(5F)=2(5F)^2

We move all terms to the left:

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

determiningTheFunctionDomain
5F-25F^2=0

We add all the numbers together, and all the variables

-25F^2+5F=0

a = -25; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·(-25)·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*-25}=\frac{-10}{-50}
=1/5
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*-25}=\frac{0}{-50}
=0
$