16t2+111t=0

Solution for 16t2+111t=0 equation:

16t^2+111t=0

a = 16; b = 111; c = 0;
Δ = b2-4ac
Δ = 1112-4·16·0
Δ = 12321
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{12321}=111$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(111)-111}{2*16}=\frac{-222}{32}
=-6+15/16
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(111)+111}{2*16}=\frac{0}{32}
=0
$