8=0.3t^2+0.1t+4.2

Solution for 8=0.3t^2+0.1t+4.2 equation:

8=0.3t^2+0.1t+4.2

We move all terms to the left:

8-(0.3t^2+0.1t+4.2)=0

We get rid of parentheses

-0.3t^2-0.1t-4.2+8=0

We add all the numbers together, and all the variables

-0.3t^2-0.1t+3.8=0

a = -0.3; b = -0.1; c = +3.8;
Δ = b2-4ac
Δ = -0.12-4·(-0.3)·3.8
Δ = 4.57
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{-(-0.1)-\sqrt{4.57}}{2*-0.3}=\frac{0.1-\sqrt{4.57}}{-0.6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.1)+\sqrt{4.57}}{2*-0.3}=\frac{0.1+\sqrt{4.57}}{-0.6}
$