0.1=t^2-18.39t

Solution for 0.1=t^2-18.39t equation:

0.1=t^2-18.39t

We move all terms to the left:

0.1-(t^2-18.39t)=0

We get rid of parentheses

-t^2+18.39t+0.1=0

We add all the numbers together, and all the variables

-1t^2+18.39t+0.1=0

a = -1; b = 18.39; c = +0.1;
Δ = b2-4ac
Δ = 18.392-4·(-1)·0.1
Δ = 338.5921
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{-(18.39)-\sqrt{338.5921}}{2*-1}=\frac{-18.39-\sqrt{338.5921}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18.39)+\sqrt{338.5921}}{2*-1}=\frac{-18.39+\sqrt{338.5921}}{-2}
$