16v^2+40v-25=0

Solution for 16v^2+40v-25=0 equation:

16v^2+40v-25=0

a = 16; b = 40; c = -25;
Δ = b2-4ac
Δ = 402-4·16·(-25)
Δ = 3200
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:

$\sqrt{\Delta}=\sqrt{3200}=\sqrt{1600*2}=\sqrt{1600}*\sqrt{2}=40\sqrt{2}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-40\sqrt{2}}{2*16}=\frac{-40-40\sqrt{2}}{32}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+40\sqrt{2}}{2*16}=\frac{-40+40\sqrt{2}}{32}
$