-3t^2+30t-22=0

Solution for -3t^2+30t-22=0 equation:

-3t^2+30t-22=0

a = -3; b = 30; c = -22;
Δ = b2-4ac
Δ = 302-4·(-3)·(-22)
Δ = 636
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:

$\sqrt{\Delta}=\sqrt{636}=\sqrt{4*159}=\sqrt{4}*\sqrt{159}=2\sqrt{159}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-2\sqrt{159}}{2*-3}=\frac{-30-2\sqrt{159}}{-6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+2\sqrt{159}}{2*-3}=\frac{-30+2\sqrt{159}}{-6}
$