f+1/4f=10

Solution for f+1/4f=10 equation:

f+1/4f=10

We move all terms to the left:

f+1/4f-(10)=0


Domain of the equation: 4f!=0

f!=0/4

f!=0

f∈R


We multiply all the terms by the denominator


f*4f-10*4f+1=0

Wy multiply elements

4f^2-40f+1=0

a = 4; b = -40; c = +1;
Δ = b2-4ac
Δ = -402-4·4·1
Δ = 1584
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:

$\sqrt{\Delta}=\sqrt{1584}=\sqrt{144*11}=\sqrt{144}*\sqrt{11}=12\sqrt{11}$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-12\sqrt{11}}{2*4}=\frac{40-12\sqrt{11}}{8}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+12\sqrt{11}}{2*4}=\frac{40+12\sqrt{11}}{8}
$