F(x)=(5+x)(x-4)

Solution for F(x)=(5+x)(x-4) equation:

(F)=(5+F)(F-4)

We move all terms to the left:

(F)-((5+F)(F-4))=0

We add all the numbers together, and all the variables

F-((F+5)(F-4))=0

We multiply parentheses ..

-((+F^2-4F+5F-20))+F=0


We calculate terms in parentheses: -((+F^2-4F+5F-20)), so:

(+F^2-4F+5F-20)

We get rid of parentheses

F^2-4F+5F-20

We add all the numbers together, and all the variables

F^2+F-20

Back to the equation:

-(F^2+F-20)


We add all the numbers together, and all the variables

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

We get rid of parentheses

-F^2+F-F+20=0

We add all the numbers together, and all the variables

-1F^2+20=0

a = -1; b = 0; c = +20;
Δ = b2-4ac
Δ = 02-4·(-1)·20
Δ = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{5}}{2*-1}=\frac{0-4\sqrt{5}}{-2}
=-\frac{4\sqrt{5}}{-2}
=-\frac{2\sqrt{5}}{-1}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{5}}{2*-1}=\frac{0+4\sqrt{5}}{-2}
=\frac{4\sqrt{5}}{-2}
=\frac{2\sqrt{5}}{-1}
$