-v^2+2v+7=0

Solution for -v^2+2v+7=0 equation:

-v^2+2v+7=0

We add all the numbers together, and all the variables

-1v^2+2v+7=0

a = -1; b = 2; c = +7;
Δ = b2-4ac
Δ = 22-4·(-1)·7
Δ = 32
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:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4\sqrt{2}}{2*-1}=\frac{-2-4\sqrt{2}}{-2}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4\sqrt{2}}{2*-1}=\frac{-2+4\sqrt{2}}{-2}
$