-v^2+2v+21=0

Solution for -v^2+2v+21=0 equation:

-v^2+2v+21=0

We add all the numbers together, and all the variables

-1v^2+2v+21=0

a = -1; b = 2; c = +21;
Δ = b2-4ac
Δ = 22-4·(-1)·21
Δ = 88
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:

$\sqrt{\Delta}=\sqrt{88}=\sqrt{4*22}=\sqrt{4}*\sqrt{22}=2\sqrt{22}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{22}}{2*-1}=\frac{-2-2\sqrt{22}}{-2}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{22}}{2*-1}=\frac{-2+2\sqrt{22}}{-2}
$