v(v+2)=13

Solution for v(v+2)=13 equation:

v(v+2)=13

We move all terms to the left:

v(v+2)-(13)=0

We multiply parentheses

v^2+2v-13=0

a = 1; b = 2; c = -13;
Δ = b2-4ac
Δ = 22-4·1·(-13)
Δ = 56
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:

$\sqrt{\Delta}=\sqrt{56}=\sqrt{4*14}=\sqrt{4}*\sqrt{14}=2\sqrt{14}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{14}}{2*1}=\frac{-2-2\sqrt{14}}{2}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{14}}{2*1}=\frac{-2+2\sqrt{14}}{2}
$