7v^2+28v-12=0

Solution for 7v^2+28v-12=0 equation:

7v^2+28v-12=0

a = 7; b = 28; c = -12;
Δ = b2-4ac
Δ = 282-4·7·(-12)
Δ = 1120
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:

$\sqrt{\Delta}=\sqrt{1120}=\sqrt{16*70}=\sqrt{16}*\sqrt{70}=4\sqrt{70}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-4\sqrt{70}}{2*7}=\frac{-28-4\sqrt{70}}{14}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+4\sqrt{70}}{2*7}=\frac{-28+4\sqrt{70}}{14}
$