6v^2+42v-54=0

Solution for 6v^2+42v-54=0 equation:

6v^2+42v-54=0

a = 6; b = 42; c = -54;
Δ = b2-4ac
Δ = 422-4·6·(-54)
Δ = 3060
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:

$\sqrt{\Delta}=\sqrt{3060}=\sqrt{36*85}=\sqrt{36}*\sqrt{85}=6\sqrt{85}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-6\sqrt{85}}{2*6}=\frac{-42-6\sqrt{85}}{12}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+6\sqrt{85}}{2*6}=\frac{-42+6\sqrt{85}}{12}
$