11t^2-14t-16=0

Solution for 11t^2-14t-16=0 equation:

11t^2-14t-16=0

a = 11; b = -14; c = -16;
Δ = b2-4ac
Δ = -142-4·11·(-16)
Δ = 900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{900}=30$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-30}{2*11}=\frac{-16}{22}
=-8/11
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+30}{2*11}=\frac{44}{22}
=2
$