6t^2-22t+9=0

Solution for 6t^2-22t+9=0 equation:

6t^2-22t+9=0

a = 6; b = -22; c = +9;
Δ = b2-4ac
Δ = -222-4·6·9
Δ = 268
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:

$\sqrt{\Delta}=\sqrt{268}=\sqrt{4*67}=\sqrt{4}*\sqrt{67}=2\sqrt{67}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{67}}{2*6}=\frac{22-2\sqrt{67}}{12}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{67}}{2*6}=\frac{22+2\sqrt{67}}{12}
$