-4-1-2v^2=-33

Solution for -4-1-2v^2=-33 equation:

-4-1-2v^2=-33

We move all terms to the left:

-4-1-2v^2-(-33)=0

We add all the numbers together, and all the variables

-2v^2+28=0

a = -2; b = 0; c = +28;
Δ = b2-4ac
Δ = 02-4·(-2)·28
Δ = 224
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:

$\sqrt{\Delta}=\sqrt{224}=\sqrt{16*14}=\sqrt{16}*\sqrt{14}=4\sqrt{14}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{14}}{2*-2}=\frac{0-4\sqrt{14}}{-4}
=-\frac{4\sqrt{14}}{-4}
=-\frac{\sqrt{14}}{-1}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{14}}{2*-2}=\frac{0+4\sqrt{14}}{-4}
=\frac{4\sqrt{14}}{-4}
=\frac{\sqrt{14}}{-1}
$