-21-6v=-v^2

Solution for -21-6v=-v^2 equation:

-21-6v=-v^2

We move all terms to the left:

-21-6v-(-v^2)=0

We get rid of parentheses

v^2-6v-21=0

a = 1; b = -6; c = -21;
Δ = b2-4ac
Δ = -62-4·1·(-21)
Δ = 120
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:

$\sqrt{\Delta}=\sqrt{120}=\sqrt{4*30}=\sqrt{4}*\sqrt{30}=2\sqrt{30}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{30}}{2*1}=\frac{6-2\sqrt{30}}{2}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{30}}{2*1}=\frac{6+2\sqrt{30}}{2}
$