6t^2+24t-18=0

Solution for 6t^2+24t-18=0 equation:

6t^2+24t-18=0

a = 6; b = 24; c = -18;
Δ = b2-4ac
Δ = 242-4·6·(-18)
Δ = 1008
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:

$\sqrt{\Delta}=\sqrt{1008}=\sqrt{144*7}=\sqrt{144}*\sqrt{7}=12\sqrt{7}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-12\sqrt{7}}{2*6}=\frac{-24-12\sqrt{7}}{12}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+12\sqrt{7}}{2*6}=\frac{-24+12\sqrt{7}}{12}
$