F(4x)=2/5x+11

Solution for F(4x)=2/5x+11 equation:

(4F)=2/5F+11

We move all terms to the left:

(4F)-(2/5F+11)=0


Domain of the equation: 5F+11)!=0

F∈R


We get rid of parentheses

4F-2/5F-11=0

We multiply all the terms by the denominator


4F*5F-11*5F-2=0

Wy multiply elements

20F^2-55F-2=0

a = 20; b = -55; c = -2;
Δ = b2-4ac
Δ = -552-4·20·(-2)
Δ = 3185
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:

$\sqrt{\Delta}=\sqrt{3185}=\sqrt{49*65}=\sqrt{49}*\sqrt{65}=7\sqrt{65}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-55)-7\sqrt{65}}{2*20}=\frac{55-7\sqrt{65}}{40}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-55)+7\sqrt{65}}{2*20}=\frac{55+7\sqrt{65}}{40}
$