3/3+d=4/d+12

Solution for 3/3+d=4/d+12 equation:

3/3+d=4/d+12

We move all terms to the left:

3/3+d-(4/d+12)=0


Domain of the equation: d+12)!=0

d∈R


We add all the numbers together, and all the variables

d-(4/d+12)+1=0

We get rid of parentheses

d-4/d-12+1=0

We multiply all the terms by the denominator


d*d-12*d+1*d-4=0

We add all the numbers together, and all the variables

-11d+d*d-4=0

Wy multiply elements

d^2-11d-4=0

a = 1; b = -11; c = -4;
Δ = b2-4ac
Δ = -112-4·1·(-4)
Δ = 137
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-\sqrt{137}}{2*1}=\frac{11-\sqrt{137}}{2}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{137}}{2*1}=\frac{11+\sqrt{137}}{2}
$