2.8t=0.04t^2+3.1826

Solution for 2.8t=0.04t^2+3.1826 equation:

2.8t=0.04t^2+3.1826

We move all terms to the left:

2.8t-(0.04t^2+3.1826)=0

We get rid of parentheses

-0.04t^2+2.8t-3.1826=0

a = -0.04; b = 2.8; c = -3.1826;
Δ = b2-4ac
Δ = 2.82-4·(-0.04)·(-3.1826)
Δ = 7.330784
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{7.330784}}{2*-0.04}=\frac{-2.8-\sqrt{7.330784}}{-0.08}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2.8)+\sqrt{7.330784}}{2*-0.04}=\frac{-2.8+\sqrt{7.330784}}{-0.08}
$