1.5t^2+3t-18=0

Solution for 1.5t^2+3t-18=0 equation:

1.5t^2+3t-18=0

a = 1.5; b = 3; c = -18;
Δ = b2-4ac
Δ = 32-4·1.5·(-18)
Δ = 117
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{117}=\sqrt{9*13}=\sqrt{9}*\sqrt{13}=3\sqrt{13}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{13}}{2*1.5}=\frac{-3-3\sqrt{13}}{3}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{13}}{2*1.5}=\frac{-3+3\sqrt{13}}{3}
$