V=4t+0.1t^2

Solution for V=4t+0.1t^2 equation:

=4V+0.1V^2

We move all terms to the left:

-(4V+0.1V^2)=0

We get rid of parentheses

-0.1V^2-4V=0

a = -0.1; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·(-0.1)·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$V_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$V_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*-0.1}=\frac{0}{-0.2}
=0
$
$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*-0.1}=\frac{8}{-0.2}
=-40
$