10-(3)/(5)f=8-(1)/(2)f

Solution for 10-(3)/(5)f=8-(1)/(2)f equation:

10-(3)/(5)f=8-(1)/(2)f

We move all terms to the left:

10-(3)/(5)f-(8-(1)/(2)f)=0


Domain of the equation: 5f!=0

f!=0/5

f!=0

f∈R



Domain of the equation: 2f)!=0

f!=0/1

f!=0

f∈R


We add all the numbers together, and all the variables

-3/5f-(-1/2f+8)+10=0

We get rid of parentheses

-3/5f+1/2f-8+10=0

We calculate fractions


(-6f)/10f^2+5f/10f^2-8+10=0

We add all the numbers together, and all the variables

(-6f)/10f^2+5f/10f^2+2=0

We multiply all the terms by the denominator


(-6f)+5f+2*10f^2=0

We add all the numbers together, and all the variables

5f+(-6f)+2*10f^2=0

Wy multiply elements

20f^2+5f+(-6f)=0

We get rid of parentheses

20f^2+5f-6f=0

We add all the numbers together, and all the variables

20f^2-1f=0

a = 20; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·20·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*20}=\frac{0}{40}
=0
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*20}=\frac{2}{40}
=1/20
$