v^2+14v=32

Solution for v^2+14v=32 equation:

v^2+14v=32

We move all terms to the left:

v^2+14v-(32)=0

a = 1; b = 14; c = -32;
Δ = b2-4ac
Δ = 142-4·1·(-32)
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{-(14)-18}{2*1}=\frac{-32}{2}
=-16
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+18}{2*1}=\frac{4}{2}
=2
$