18t^2-5t-2=0

Solution for 18t^2-5t-2=0 equation:

18t^2-5t-2=0

a = 18; b = -5; c = -2;
Δ = b2-4ac
Δ = -52-4·18·(-2)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-13}{2*18}=\frac{-8}{36}
=-2/9
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+13}{2*18}=\frac{18}{36}
=1/2
$