3t2+6t-18=0

Solution for 3t2+6t-18=0 equation:

3t^2+6t-18=0

a = 3; b = 6; c = -18;
Δ = b2-4ac
Δ = 62-4·3·(-18)
Δ = 252
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:

$\sqrt{\Delta}=\sqrt{252}=\sqrt{36*7}=\sqrt{36}*\sqrt{7}=6\sqrt{7}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6\sqrt{7}}{2*3}=\frac{-6-6\sqrt{7}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{7}}{2*3}=\frac{-6+6\sqrt{7}}{6}
$