(2b)^2+b^2=100

Solution for (2b)^2+b^2=100 equation:

(2b)^2+b^2=100

We move all terms to the left:

(2b)^2+b^2-(100)=0

We add all the numbers together, and all the variables

3b^2-100=0

a = 3; b = 0; c = -100;
Δ = b2-4ac
Δ = 02-4·3·(-100)
Δ = 1200
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:

$\sqrt{\Delta}=\sqrt{1200}=\sqrt{400*3}=\sqrt{400}*\sqrt{3}=20\sqrt{3}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{3}}{2*3}=\frac{0-20\sqrt{3}}{6}
=-\frac{20\sqrt{3}}{6}
=-\frac{10\sqrt{3}}{3}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{3}}{2*3}=\frac{0+20\sqrt{3}}{6}
=\frac{20\sqrt{3}}{6}
=\frac{10\sqrt{3}}{3}
$