v^2+6v-68=4

Solution for v^2+6v-68=4 equation:

v^2+6v-68=4

We move all terms to the left:

v^2+6v-68-(4)=0

We add all the numbers together, and all the variables

v^2+6v-72=0

a = 1; b = 6; c = -72;
Δ = b2-4ac
Δ = 62-4·1·(-72)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-18}{2*1}=\frac{-24}{2}
=-12
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+18}{2*1}=\frac{12}{2}
=6
$