20-(1/5d)=3/10d+16

Solution for 20-(1/5d)=3/10d+16 equation:

20-(1/5d)=3/10d+16

We move all terms to the left:

20-(1/5d)-(3/10d+16)=0


Domain of the equation: 5d)!=0

d!=0/1

d!=0

d∈R



Domain of the equation: 10d+16)!=0

d∈R


We add all the numbers together, and all the variables

-(+1/5d)-(3/10d+16)+20=0

We get rid of parentheses

-1/5d-3/10d-16+20=0

We calculate fractions


(-10d)/50d^2+(-15d)/50d^2-16+20=0

We add all the numbers together, and all the variables

(-10d)/50d^2+(-15d)/50d^2+4=0

We multiply all the terms by the denominator


(-10d)+(-15d)+4*50d^2=0

Wy multiply elements

200d^2+(-10d)+(-15d)=0

We get rid of parentheses

200d^2-10d-15d=0

We add all the numbers together, and all the variables

200d^2-25d=0

a = 200; b = -25; c = 0;
Δ = b2-4ac
Δ = -252-4·200·0
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-25}{2*200}=\frac{0}{400}
=0
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+25}{2*200}=\frac{50}{400}
=1/8
$