3t^2-6t-17=0

Solution for 3t^2-6t-17=0 equation:

3t^2-6t-17=0

a = 3; b = -6; c = -17;
Δ = b2-4ac
Δ = -62-4·3·(-17)
Δ = 240
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:

$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-4\sqrt{15}}{2*3}=\frac{6-4\sqrt{15}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+4\sqrt{15}}{2*3}=\frac{6+4\sqrt{15}}{6}
$