1/2f+3=13-f

Solution for 1/2f+3=13-f equation:

1/2f+3=13-f

We move all terms to the left:

1/2f+3-(13-f)=0


Domain of the equation: 2f!=0

f!=0/2

f!=0

f∈R


We add all the numbers together, and all the variables

1/2f-(-1f+13)+3=0

We get rid of parentheses

1/2f+1f-13+3=0

We multiply all the terms by the denominator


1f*2f-13*2f+3*2f+1=0

Wy multiply elements

2f^2-26f+6f+1=0

We add all the numbers together, and all the variables

2f^2-20f+1=0

a = 2; b = -20; c = +1;
Δ = b2-4ac
Δ = -202-4·2·1
Δ = 392
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:

$\sqrt{\Delta}=\sqrt{392}=\sqrt{196*2}=\sqrt{196}*\sqrt{2}=14\sqrt{2}$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-14\sqrt{2}}{2*2}=\frac{20-14\sqrt{2}}{4}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+14\sqrt{2}}{2*2}=\frac{20+14\sqrt{2}}{4}
$