3t^2-24t+22=0

Solution for 3t^2-24t+22=0 equation:

3t^2-24t+22=0

a = 3; b = -24; c = +22;
Δ = b2-4ac
Δ = -242-4·3·22
Δ = 312
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:

$\sqrt{\Delta}=\sqrt{312}=\sqrt{4*78}=\sqrt{4}*\sqrt{78}=2\sqrt{78}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-2\sqrt{78}}{2*3}=\frac{24-2\sqrt{78}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+2\sqrt{78}}{2*3}=\frac{24+2\sqrt{78}}{6}
$