2(10b+4b)b=2

Solution for 2(10b+4b)b=2 equation:

2(10b+4b)b=2

We move all terms to the left:

2(10b+4b)b-(2)=0

We add all the numbers together, and all the variables

2(+14b)b-2=0

We multiply parentheses

28b^2-2=0

a = 28; b = 0; c = -2;
Δ = b2-4ac
Δ = 02-4·28·(-2)
Δ = 224
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:

$\sqrt{\Delta}=\sqrt{224}=\sqrt{16*14}=\sqrt{16}*\sqrt{14}=4\sqrt{14}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{14}}{2*28}=\frac{0-4\sqrt{14}}{56}
=-\frac{4\sqrt{14}}{56}
=-\frac{\sqrt{14}}{14}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{14}}{2*28}=\frac{0+4\sqrt{14}}{56}
=\frac{4\sqrt{14}}{56}
=\frac{\sqrt{14}}{14}
$