A2+7a2=128

Solution for A2+7a2=128 equation:

2+7A^2=128

We move all terms to the left:

2+7A^2-(128)=0

We add all the numbers together, and all the variables

7A^2-126=0

a = 7; b = 0; c = -126;
Δ = b2-4ac
Δ = 02-4·7·(-126)
Δ = 3528
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{3528}=\sqrt{1764*2}=\sqrt{1764}*\sqrt{2}=42\sqrt{2}$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-42\sqrt{2}}{2*7}=\frac{0-42\sqrt{2}}{14}
=-\frac{42\sqrt{2}}{14}
=-3\sqrt{2}
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+42\sqrt{2}}{2*7}=\frac{0+42\sqrt{2}}{14}
=\frac{42\sqrt{2}}{14}
=3\sqrt{2}
$