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

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
$