8v(v+2)=0

Solution for 8v(v+2)=0 equation:

8v(v+2)=0

We multiply parentheses

8v^2+16v=0

a = 8; b = 16; c = 0;
Δ = b2-4ac
Δ = 162-4·8·0
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16}{2*8}=\frac{-32}{16}
=-2
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16}{2*8}=\frac{0}{16}
=0
$