D=(3x-2)(7x+1)-2(3x-2)

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Solution for D=(3x-2)(7x+1)-2(3x-2) equation:



=(3D-2)(7D+1)-2(3D-2)
We move all terms to the left:
-((3D-2)(7D+1)-2(3D-2))=0
We multiply parentheses ..
-((+21D^2+3D-14D-2)-2(3D-2))=0
We calculate terms in parentheses: -((+21D^2+3D-14D-2)-2(3D-2)), so:
(+21D^2+3D-14D-2)-2(3D-2)
We multiply parentheses
(+21D^2+3D-14D-2)-6D+4
We get rid of parentheses
21D^2+3D-14D-6D-2+4
We add all the numbers together, and all the variables
21D^2-17D+2
Back to the equation:
-(21D^2-17D+2)
We get rid of parentheses
-21D^2+17D-2=0
a = -21; b = 17; c = -2;
Δ = b2-4ac
Δ = 172-4·(-21)·(-2)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-11}{2*-21}=\frac{-28}{-42} =2/3 $
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+11}{2*-21}=\frac{-6}{-42} =1/7 $

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