12t^2-15t-18=0

Solution for 12t^2-15t-18=0 equation:

12t^2-15t-18=0

a = 12; b = -15; c = -18;
Δ = b2-4ac
Δ = -152-4·12·(-18)
Δ = 1089
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1089}=33$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-33}{2*12}=\frac{-18}{24}
=-3/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+33}{2*12}=\frac{48}{24}
=2
$