16(4)=v^2

Solution for 16(4)=v^2 equation:

16(4)=v^2

We move all terms to the left:

16(4)-(v^2)=0

We add all the numbers together, and all the variables

-1v^2+164=0

a = -1; b = 0; c = +164;
Δ = b2-4ac
Δ = 02-4·(-1)·164
Δ = 656
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{656}=\sqrt{16*41}=\sqrt{16}*\sqrt{41}=4\sqrt{41}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{41}}{2*-1}=\frac{0-4\sqrt{41}}{-2}
=-\frac{4\sqrt{41}}{-2}
=-\frac{2\sqrt{41}}{-1}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{41}}{2*-1}=\frac{0+4\sqrt{41}}{-2}
=\frac{4\sqrt{41}}{-2}
=\frac{2\sqrt{41}}{-1}
$