I(t)=-0.1t^2+1.2t

Solution for I(t)=-0.1t^2+1.2t equation:

(I)=-0.1I^2+1.2I

We move all terms to the left:

(I)-(-0.1I^2+1.2I)=0

We get rid of parentheses

0.1I^2-1.2I+I=0

We add all the numbers together, and all the variables

0.1I^2-0.2I=0

a = 0.1; b = -0.2; c = 0;
Δ = b2-4ac
Δ = -0.22-4·0.1·0
Δ = 0.04
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$I_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$I_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$I_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-0.2)-\sqrt{0.04}}{2*0.1}=\frac{0.2-\sqrt{0.04}}{0.2}
$
$I_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.2)+\sqrt{0.04}}{2*0.1}=\frac{0.2+\sqrt{0.04}}{0.2}
$