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

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

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

We move all terms to the left:

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

We get rid of parentheses

0.1T^2-1.2T+T=0

We add all the numbers together, and all the variables

0.1T^2-0.2T=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:
$T_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$T_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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