-0.1t^2+2.8t+2=10

Solution for -0.1t^2+2.8t+2=10 equation:

-0.1t^2+2.8t+2=10

We move all terms to the left:

-0.1t^2+2.8t+2-(10)=0

We add all the numbers together, and all the variables

-0.1t^2+2.8t-8=0

a = -0.1; b = 2.8; c = -8;
Δ = b2-4ac
Δ = 2.82-4·(-0.1)·(-8)
Δ = 4.64
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{-(2.8)-\sqrt{4.64}}{2*-0.1}=\frac{-2.8-\sqrt{4.64}}{-0.2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2.8)+\sqrt{4.64}}{2*-0.1}=\frac{-2.8+\sqrt{4.64}}{-0.2}
$