2t^2-11t-22=0

Solution for 2t^2-11t-22=0 equation:

2t^2-11t-22=0

a = 2; b = -11; c = -22;
Δ = b2-4ac
Δ = -112-4·2·(-22)
Δ = 297
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:

$\sqrt{\Delta}=\sqrt{297}=\sqrt{9*33}=\sqrt{9}*\sqrt{33}=3\sqrt{33}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-3\sqrt{33}}{2*2}=\frac{11-3\sqrt{33}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+3\sqrt{33}}{2*2}=\frac{11+3\sqrt{33}}{4}
$