6t^2+33t-12=0

Solution for 6t^2+33t-12=0 equation:

6t^2+33t-12=0

a = 6; b = 33; c = -12;
Δ = b2-4ac
Δ = 332-4·6·(-12)
Δ = 1377
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:

$\sqrt{\Delta}=\sqrt{1377}=\sqrt{81*17}=\sqrt{81}*\sqrt{17}=9\sqrt{17}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-9\sqrt{17}}{2*6}=\frac{-33-9\sqrt{17}}{12}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+9\sqrt{17}}{2*6}=\frac{-33+9\sqrt{17}}{12}
$