16=1/2*4(3*b2)

Solution for 16=1/2*4(3*b2) equation:

16=1/2*4(3b^2)

We move all terms to the left:

16-(1/2*4(3b^2))=0


Domain of the equation: 2*43b^2)!=0

b!=0/1

b!=0

b∈R


We get rid of parentheses

-1/2*43b^2+16=0

We multiply all the terms by the denominator


16*2*43b^2-1=0

Wy multiply elements

1376b^2*4-1=0

Wy multiply elements

5504b^2-1=0

a = 5504; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·5504·(-1)
Δ = 22016
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:

$\sqrt{\Delta}=\sqrt{22016}=\sqrt{256*86}=\sqrt{256}*\sqrt{86}=16\sqrt{86}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{86}}{2*5504}=\frac{0-16\sqrt{86}}{11008}
=-\frac{16\sqrt{86}}{11008}
=-\frac{\sqrt{86}}{688}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{86}}{2*5504}=\frac{0+16\sqrt{86}}{11008}
=\frac{16\sqrt{86}}{11008}
=\frac{\sqrt{86}}{688}
$