(2b)^2+b^2=196

Solution for (2b)^2+b^2=196 equation:

(2b)^2+b^2=196

We move all terms to the left:

(2b)^2+b^2-(196)=0

We add all the numbers together, and all the variables

3b^2-196=0

a = 3; b = 0; c = -196;
Δ = b2-4ac
Δ = 02-4·3·(-196)
Δ = 2352
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2352}=\sqrt{784*3}=\sqrt{784}*\sqrt{3}=28\sqrt{3}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-28\sqrt{3}}{2*3}=\frac{0-28\sqrt{3}}{6}
=-\frac{28\sqrt{3}}{6}
=-\frac{14\sqrt{3}}{3}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+28\sqrt{3}}{2*3}=\frac{0+28\sqrt{3}}{6}
=\frac{28\sqrt{3}}{6}
=\frac{14\sqrt{3}}{3}
$