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

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

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

We move all terms to the left:

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


Domain of the equation: d!=0

d∈R



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

d∈R


We get rid of parentheses

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

We calculate fractions


(-10d)/10d^2+(-3d)/10d^2-16+20=0

We add all the numbers together, and all the variables

(-10d)/10d^2+(-3d)/10d^2+4=0

We multiply all the terms by the denominator


(-10d)+(-3d)+4*10d^2=0

Wy multiply elements

40d^2+(-10d)+(-3d)=0

We get rid of parentheses

40d^2-10d-3d=0

We add all the numbers together, and all the variables

40d^2-13d=0

a = 40; b = -13; c = 0;
Δ = b2-4ac
Δ = -132-4·40·0
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-13}{2*40}=\frac{0}{80}
=0
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+13}{2*40}=\frac{26}{80}
=13/40
$