(1/3)f+1=5

Solution for (1/3)f+1=5 equation:

(1/3)f+1=5

We move all terms to the left:

(1/3)f+1-(5)=0


Domain of the equation: 3)f!=0

f!=0/1

f!=0

f∈R


We add all the numbers together, and all the variables

(+1/3)f+1-5=0

We add all the numbers together, and all the variables

(+1/3)f-4=0

We multiply parentheses

f^2-4=0

a = 1; b = 0; c = -4;
Δ = b2-4ac
Δ = 02-4·1·(-4)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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