4v^2+27v-18=0

Solution for 4v^2+27v-18=0 equation:

4v^2+27v-18=0

a = 4; b = 27; c = -18;
Δ = b2-4ac
Δ = 272-4·4·(-18)
Δ = 1017
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:

$\sqrt{\Delta}=\sqrt{1017}=\sqrt{9*113}=\sqrt{9}*\sqrt{113}=3\sqrt{113}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-3\sqrt{113}}{2*4}=\frac{-27-3\sqrt{113}}{8}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+3\sqrt{113}}{2*4}=\frac{-27+3\sqrt{113}}{8}
$