2/5f-3=0.6f+10

Solution for 2/5f-3=0.6f+10 equation:

2/5f-3=0.6f+10

We move all terms to the left:

2/5f-3-(0.6f+10)=0


Domain of the equation: 5f!=0

f!=0/5

f!=0

f∈R


We get rid of parentheses

2/5f-0.6f-10-3=0

We multiply all the terms by the denominator


-(0.6f)*5f-10*5f-3*5f+2=0

We add all the numbers together, and all the variables

-(+0.6f)*5f-10*5f-3*5f+2=0

We multiply parentheses

-0f^2-10*5f-3*5f+2=0

Wy multiply elements

-0f^2-50f-15f+2=0

We add all the numbers together, and all the variables

-1f^2-65f+2=0

a = -1; b = -65; c = +2;
Δ = b2-4ac
Δ = -652-4·(-1)·2
Δ = 4233
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-65)-\sqrt{4233}}{2*-1}=\frac{65-\sqrt{4233}}{-2}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-65)+\sqrt{4233}}{2*-1}=\frac{65+\sqrt{4233}}{-2}
$