-4/10f-2/5f=40

Solution for -4/10f-2/5f=40 equation:

-4/10f-2/5f=40

We move all terms to the left:

-4/10f-2/5f-(40)=0


Domain of the equation: 10f!=0

f!=0/10

f!=0

f∈R



Domain of the equation: 5f!=0

f!=0/5

f!=0

f∈R


We calculate fractions


(-20f)/50f^2+(-20f)/50f^2-40=0

We multiply all the terms by the denominator


(-20f)+(-20f)-40*50f^2=0

Wy multiply elements

-2000f^2+(-20f)+(-20f)=0

We get rid of parentheses

-2000f^2-20f-20f=0

We add all the numbers together, and all the variables

-2000f^2-40f=0

a = -2000; b = -40; c = 0;
Δ = b2-4ac
Δ = -402-4·(-2000)·0
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-40}{2*-2000}=\frac{0}{-4000}
=0
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+40}{2*-2000}=\frac{80}{-4000}
=-1/50
$