V=2t^2+4t-2

Solution for V=2t^2+4t-2 equation:

=2V^2+4V-2

We move all terms to the left:

-(2V^2+4V-2)=0

We get rid of parentheses

-2V^2-4V+2=0

a = -2; b = -4; c = +2;
Δ = b2-4ac
Δ = -42-4·(-2)·2
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{-(-4)-4\sqrt{2}}{2*-2}=\frac{4-4\sqrt{2}}{-4}
$
$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{2}}{2*-2}=\frac{4+4\sqrt{2}}{-4}
$