18t^2+27t-26=0

Solution for 18t^2+27t-26=0 equation:

18t^2+27t-26=0

a = 18; b = 27; c = -26;
Δ = b2-4ac
Δ = 272-4·18·(-26)
Δ = 2601
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2601}=51$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-51}{2*18}=\frac{-78}{36}
=-2+1/6
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+51}{2*18}=\frac{24}{36}
=2/3
$