A=(10-7z)(10+7z)

Solution for A=(10-7z)(10+7z) equation:

=(10-7A)(10+7A)

We move all terms to the left:

-((10-7A)(10+7A))=0

We add all the numbers together, and all the variables

-((-7A+10)(7A+10))=0

We multiply parentheses ..

-((-49A^2-70A+70A+100))=0


We calculate terms in parentheses: -((-49A^2-70A+70A+100)), so:

(-49A^2-70A+70A+100)

We get rid of parentheses

-49A^2-70A+70A+100

We add all the numbers together, and all the variables

-49A^2+100

Back to the equation:

-(-49A^2+100)


We get rid of parentheses

49A^2-100=0

a = 49; b = 0; c = -100;
Δ = b2-4ac
Δ = 02-4·49·(-100)
Δ = 19600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{19600}=140$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-140}{2*49}=\frac{-140}{98}
=-1+3/7
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+140}{2*49}=\frac{140}{98}
=1+3/7
$