V(t)=2t^2+t-6

Solution for V(t)=2t^2+t-6 equation:

(V)=2V^2+V-6

We move all terms to the left:

(V)-(2V^2+V-6)=0

We get rid of parentheses

-2V^2+V-V+6=0

We add all the numbers together, and all the variables

-2V^2+6=0

a = -2; b = 0; c = +6;
Δ = b2-4ac
Δ = 02-4·(-2)·6
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$V_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{3}}{2*-2}=\frac{0-4\sqrt{3}}{-4}
=-\frac{4\sqrt{3}}{-4}
=-\frac{\sqrt{3}}{-1}
$
$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{3}}{2*-2}=\frac{0+4\sqrt{3}}{-4}
=\frac{4\sqrt{3}}{-4}
=\frac{\sqrt{3}}{-1}
$