1/3d+24=5/9d-12

Solution for 1/3d+24=5/9d-12 equation:

1/3d+24=5/9d-12

We move all terms to the left:

1/3d+24-(5/9d-12)=0


Domain of the equation: 3d!=0

d!=0/3

d!=0

d∈R



Domain of the equation: 9d-12)!=0

d∈R


We get rid of parentheses

1/3d-5/9d+12+24=0

We calculate fractions


9d/27d^2+(-15d)/27d^2+12+24=0

We add all the numbers together, and all the variables

9d/27d^2+(-15d)/27d^2+36=0

We multiply all the terms by the denominator


9d+(-15d)+36*27d^2=0

Wy multiply elements

972d^2+9d+(-15d)=0

We get rid of parentheses

972d^2+9d-15d=0

We add all the numbers together, and all the variables

972d^2-6d=0

a = 972; b = -6; c = 0;
Δ = b2-4ac
Δ = -62-4·972·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6}{2*972}=\frac{0}{1944}
=0
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6}{2*972}=\frac{12}{1944}
=1/162
$