8v^2+25v+3=0

Solution for 8v^2+25v+3=0 equation:

8v^2+25v+3=0

a = 8; b = 25; c = +3;
Δ = b2-4ac
Δ = 252-4·8·3
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-23}{2*8}=\frac{-48}{16}
=-3
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+23}{2*8}=\frac{-2}{16}
=-1/8
$