I(t)=-0.1t^2+2t

Solution for I(t)=-0.1t^2+2t equation:

(I)=-0.1I^2+2I

We move all terms to the left:

(I)-(-0.1I^2+2I)=0

We get rid of parentheses

0.1I^2-2I+I=0

We add all the numbers together, and all the variables

0.1I^2-1I=0

a = 0.1; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·0.1·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$I_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$I_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$I_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*0.1}=\frac{0}{0.2}
=0
$
$I_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*0.1}=\frac{2}{0.2}
=10
$