18t^2+39t=-20

Solution for 18t^2+39t=-20 equation:

18t^2+39t=-20

We move all terms to the left:

18t^2+39t-(-20)=0

We add all the numbers together, and all the variables

18t^2+39t+20=0

a = 18; b = 39; c = +20;
Δ = b2-4ac
Δ = 392-4·18·20
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(39)-9}{2*18}=\frac{-48}{36}
=-1+1/3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+9}{2*18}=\frac{-30}{36}
=-5/6
$