-24v^2+12v+1=0

Solution for -24v^2+12v+1=0 equation:

-24v^2+12v+1=0

a = -24; b = 12; c = +1;
Δ = b2-4ac
Δ = 122-4·(-24)·1
Δ = 240
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:

$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{15}}{2*-24}=\frac{-12-4\sqrt{15}}{-48}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{15}}{2*-24}=\frac{-12+4\sqrt{15}}{-48}
$