15t^2=18t

Solution for 15t^2=18t equation:

15t^2=18t

We move all terms to the left:

15t^2-(18t)=0

a = 15; b = -18; c = 0;
Δ = b2-4ac
Δ = -182-4·15·0
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-18}{2*15}=\frac{0}{30}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+18}{2*15}=\frac{36}{30}
=1+1/5
$