3t^2+12t-33=0

Solution for 3t^2+12t-33=0 equation:

3t^2+12t-33=0

a = 3; b = 12; c = -33;
Δ = b2-4ac
Δ = 122-4·3·(-33)
Δ = 540
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:

$\sqrt{\Delta}=\sqrt{540}=\sqrt{36*15}=\sqrt{36}*\sqrt{15}=6\sqrt{15}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-6\sqrt{15}}{2*3}=\frac{-12-6\sqrt{15}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+6\sqrt{15}}{2*3}=\frac{-12+6\sqrt{15}}{6}
$