24v^2+42v-27=0

Solution for 24v^2+42v-27=0 equation:

24v^2+42v-27=0

a = 24; b = 42; c = -27;
Δ = b2-4ac
Δ = 422-4·24·(-27)
Δ = 4356
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4356}=66$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-66}{2*24}=\frac{-108}{48}
=-2+1/4
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+66}{2*24}=\frac{24}{48}
=1/2
$