(4d-2)(3d-2)=d

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Solution for (4d-2)(3d-2)=d equation:



(4d-2)(3d-2)=d
We move all terms to the left:
(4d-2)(3d-2)-(d)=0
We add all the numbers together, and all the variables
-1d+(4d-2)(3d-2)=0
We multiply parentheses ..
(+12d^2-8d-6d+4)-1d=0
We get rid of parentheses
12d^2-8d-6d-1d+4=0
We add all the numbers together, and all the variables
12d^2-15d+4=0
a = 12; b = -15; c = +4;
Δ = b2-4ac
Δ = -152-4·12·4
Δ = 33
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{-(-15)-\sqrt{33}}{2*12}=\frac{15-\sqrt{33}}{24} $
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{33}}{2*12}=\frac{15+\sqrt{33}}{24} $

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