F(20)=(x+0.25x)(50-x)

Solution for F(20)=(x+0.25x)(50-x) equation:

(20)=(F+0.25F)(50-F)

We move all terms to the left:

(20)-((F+0.25F)(50-F))=0

We add all the numbers together, and all the variables

-((+1.25F)(-1F+50))+20=0

We multiply parentheses ..

-((-1F^2+50F))+20=0


We calculate terms in parentheses: -((-1F^2+50F)), so:

(-1F^2+50F)

We get rid of parentheses

-1F^2+50F

Back to the equation:

-(-1F^2+50F)


We get rid of parentheses

1F^2-50F+20=0

We add all the numbers together, and all the variables

F^2-50F+20=0

a = 1; b = -50; c = +20;
Δ = b2-4ac
Δ = -502-4·1·20
Δ = 2420
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2420}=\sqrt{484*5}=\sqrt{484}*\sqrt{5}=22\sqrt{5}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-22\sqrt{5}}{2*1}=\frac{50-22\sqrt{5}}{2}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+22\sqrt{5}}{2*1}=\frac{50+22\sqrt{5}}{2}
$