16v^2+18v=0

Solution for 16v^2+18v=0 equation:

16v^2+18v=0

a = 16; b = 18; c = 0;
Δ = b2-4ac
Δ = 182-4·16·0
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-18}{2*16}=\frac{-36}{32}
=-1+1/8
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+18}{2*16}=\frac{0}{32}
=0
$