-3v^2=-240

Solution for -3v^2=-240 equation:

-3v^2=-240

We move all terms to the left:

-3v^2-(-240)=0

We add all the numbers together, and all the variables

-3v^2+240=0

a = -3; b = 0; c = +240;
Δ = b2-4ac
Δ = 02-4·(-3)·240
Δ = 2880
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:

$\sqrt{\Delta}=\sqrt{2880}=\sqrt{576*5}=\sqrt{576}*\sqrt{5}=24\sqrt{5}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24\sqrt{5}}{2*-3}=\frac{0-24\sqrt{5}}{-6}
=-\frac{24\sqrt{5}}{-6}
=-\frac{4\sqrt{5}}{-1}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24\sqrt{5}}{2*-3}=\frac{0+24\sqrt{5}}{-6}
=\frac{24\sqrt{5}}{-6}
=\frac{4\sqrt{5}}{-1}
$