F(r)=(r-3)(2r-8)

Solution for F(r)=(r-3)(2r-8) equation:

(F)=(F-3)(2F-8)

We move all terms to the left:

(F)-((F-3)(2F-8))=0

We multiply parentheses ..

-((+2F^2-8F-6F+24))+F=0


We calculate terms in parentheses: -((+2F^2-8F-6F+24)), so:

(+2F^2-8F-6F+24)

We get rid of parentheses

2F^2-8F-6F+24

We add all the numbers together, and all the variables

2F^2-14F+24

Back to the equation:

-(2F^2-14F+24)


We add all the numbers together, and all the variables

F-(2F^2-14F+24)=0

We get rid of parentheses

-2F^2+F+14F-24=0

We add all the numbers together, and all the variables

-2F^2+15F-24=0

a = -2; b = 15; c = -24;
Δ = b2-4ac
Δ = 152-4·(-2)·(-24)
Δ = 33
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-\sqrt{33}}{2*-2}=\frac{-15-\sqrt{33}}{-4}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{33}}{2*-2}=\frac{-15+\sqrt{33}}{-4}
$