A=21x^2+3x

Solution for A=21x^2+3x equation:

=21A^2+3A

We move all terms to the left:

-(21A^2+3A)=0

We get rid of parentheses

-21A^2-3A=0

a = -21; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·(-21)·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*-21}=\frac{0}{-42}
=0
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*-21}=\frac{6}{-42}
=-1/7
$