10v+39=v^2

Solution for 10v+39=v^2 equation:

10v+39=v^2

We move all terms to the left:

10v+39-(v^2)=0

determiningTheFunctionDomain
-v^2+10v+39=0

We add all the numbers together, and all the variables

-1v^2+10v+39=0

a = -1; b = 10; c = +39;
Δ = b2-4ac
Δ = 102-4·(-1)·39
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-16}{2*-1}=\frac{-26}{-2}
=+13
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+16}{2*-1}=\frac{6}{-2}
=-3
$