t=0.18t^2

Solution for t=0.18t^2 equation:

t=0.18t^2

We move all terms to the left:

t-(0.18t^2)=0

We get rid of parentheses

-0.18t^2+t=0

a = -0.18; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-0.18)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*-0.18}=\frac{-2}{-0.36}
=5+0.2/0.36
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-0.18}=\frac{0}{-0.36}
=0
$