1.65t^2+6t-12=0

Solution for 1.65t^2+6t-12=0 equation:

1.65t^2+6t-12=0

a = 1.65; b = 6; c = -12;
Δ = b2-4ac
Δ = 62-4·1.65·(-12)
Δ = 115.2
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-\sqrt{115.2}}{2*1.65}=\frac{-6-\sqrt{115.2}}{3.3}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+\sqrt{115.2}}{2*1.65}=\frac{-6+\sqrt{115.2}}{3.3}
$